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Let \\(r:X\\to Y\\) be a representation and let \\(f:X\\to Z\\) be the question we want to answer. The representation is sufficient for this question precisely when there is a function \\(\\bar f:r(X)\\to Z\\) such that\n\\[\nf=\\bar f\\circ r,\n\\]\nwhere the codomain of \\(r\\) is restricted to its image in the composition. Equivalently, whenever \\(r(x)=r(x')\\), we must have \\(f(x)=f(x')\\). Necessity follows by applying \\(\\bar f\\). For sufficiency, define \\(\\bar f(r(x))=f(x)\\). Constancy on the fibers makes this definition unambiguous.","dependence has a simple formulation. Let \\(r:X\\to Y\\) be a representation and let \\(f:X\\to Z\\) be the question we want to answer. The representation is sufficient for this question precisely when there is a function \\(\\bar f:r(X)\\to Z\\) such that\n\\[\nf=\\bar f\\circ r,\n\\]\nwhere the codomain of \\(r\\) is restricted to its image in the composition. Equivalently, whenever \\(r(x)=r(x')\\), we must have \\(f(x)=f(x')\\). Necessity follows by applying \\(\\bar f\\). For sufficiency, define \\(\\bar f(r(x))=f(x)\\). Constancy on the fibers makes this definition unambiguous.","X\\to Y\\) be a representation and let \\(f:X\\to Z\\) be the question we want to answer. The representation is sufficient for this question precisely when there is a function \\(\\bar f:r(X)\\to Z\\) such that\n\\[\nf=\\bar f\\circ r,\n\\]\nwhere the codomain of \\(r\\) is restricted to its image in the composition. Equivalently, whenever \\(r(x)=r(x')\\), we must have \\(f(x)=f(x')\\). Necessity follows by applying \\(\\bar f\\). For sufficiency, define \\(\\bar f(r(x))=f(x)\\). Constancy on the fibers makes this definition unambiguous.","\\(f:X\\to Z\\) be the question we want to answer. The representation is sufficient for this question precisely when there is a function \\(\\bar f:r(X)\\to Z\\) such that\n\\[\nf=\\bar f\\circ r,\n\\]\nwhere the codomain of \\(r\\) is restricted to its image in the composition. Equivalently, whenever \\(r(x)=r(x')\\), we must have \\(f(x)=f(x')\\). Necessity follows by applying \\(\\bar f\\). For sufficiency, define \\(\\bar f(r(x))=f(x)\\). Constancy on the fibers makes this definition unambiguous.","The criterion does not require \\(r\\) to preserve everything about \\(x\\). It requires preservation of the distinctions that matter to \\(f\\). Parity is enough to decide whether an integer is even, but not enough to recover the integer. A failure of recovery is therefore a failure relative to a particular question. Conversely, a representation adequate for today's question may be inadequate when tomorrow's question asks for more.","Continuation makes the requirement sharper. For a transformation \\(T:X\\to X\\), a closed evolution \\(\\bar T\\) on the represented states must satisfy \\(rT=\\bar T r\\). Applying the same factorization criterion to \\(rT\\) shows exactly what is needed: whenever two states have the same representation, their successors must also have the same representation. Induction then gives \\(rT^n=\\bar T^n r\\) for every \\(n\\). Agreement at one observation is weaker than preservation under every continuation. This criterion concerns every state in the declared domain. Restricting the initial states changes the question.","ation makes the requirement sharper. For a transformation \\(T:X\\to X\\), a closed evolution \\(\\bar T\\) on the represented states must satisfy \\(rT=\\bar T r\\). Applying the same factorization criterion to \\(rT\\) shows exactly what is needed: whenever two states have the same representation, their successors must also have the same representation. Induction then gives \\(rT^n=\\bar T^n r\\) for every \\(n\\). Agreement at one observation is weaker than preservation under every continuation. This criterion concerns every state in the declared domain. Restricting the initial states changes the question.","An elementary tower shows why this matters. For each natural number \\(n\\), let\n\\[\nX_n=\\{m\\in\\mathbb N:m\\geq n\\},\n\\]\nand let \\(X_{n+1}\\to X_n\\) be inclusion. For any finite prefix through \\(N\\), the constant value \\(N\\) gives a compatible choice at every level in that prefix. No finite stage defeats the construction.","A global compatible family would be different. Because every restriction is inclusion, all its entries would have to be the same natural number \\(m\\). Membership at every level would require \\(m\\geq n\\) for every \\(n\\). Taking \\(n=m+1\\) gives a contradiction. Every finite prefix has a witness, but the tower has no global point.","The finite witnesses have not secretly assembled into an infinite witness. The choice for prefix \\(N+1\\) does not restrict to the previously selected constant choice \\(N\\). This is precisely the missing compatibility. There is no conflict with compactness principles that impose additional hypotheses. The sets and maps just given do not supply those hypotheses. The example identifies an obligation rather than a paradox.","Let \\(P\\) be a projector on a vector space, so \\(P^2=P\\), and put \\(Q=I-P\\). The image of \\(P\\) is the retained space. The image of \\(Q\\) is its complementary space. No inner product is required. If \\(T\\) is a linear operator, the compressed operator \\(PTP\\) records only the part that begins and ends in the retained space.","Insert \\(I=P+Q\\) between two applications of \\(T\\). Exact algebra gives\n\\begin{equation}\\label{eq:return}\nPT^2P=(PTP)^2+PTQTP.\n\\end{equation}\nThe second term has a plain meaning. The input leaves the retained space through \\(QTP\\) and later returns through \\(PTQ\\). Deleting that term is not a harmless change of notation. It changes the evolution unless a further condition makes the return vanish.","Work over \\(\\mathbb Q^3\\), with\n\\[\nT=\\begin{pmatrix}2&1&0","Work over \\(\\mathbb Q^3\\), with\n\\[\nT=\\begin{pmatrix}2&1&0\\\\0&3&1\\\\1&-1&2\\end{pmatrix},\n\\qquad P=\\operatorname{diag}(1,1,0),\n\\qquad x=\\begin{pmatrix}1\\\\2\\\\0\\end{pmatrix}.\n\\]\nHere \\(Tx=(4,6,-1)^{\\mathsf T}\\), so both descriptions give the retained observation \\((4,6,0)^{\\mathsf T}\\). This agreement follows from \\(Px=x\\), not from closure under repetition.","0\\end{pmatrix}.\n\\]\nHere \\(Tx=(4,6,-1)^{\\mathsf T}\\), so both descriptions give the retained observation \\((4,6,0)^{\\mathsf T}\\). This agreement follows from \\(Px=x\\), not from closure under repetition.","After two steps the comparison changes:\n\\[\nPT^2x=\\begin{pmatrix}14\\\\17\\\\0\\end{pmatrix},\n\\qquad\n(PTP)^2x=\\begin{pmatrix}14\\\\18\\\\0\\end{pmatrix}.\n\\]\nThe difference \\(PTQTPx=(0,-1,0)^{\\mathsf T}\\) is the return of the discarded coordinate. Equation~\\eqref{eq:return} predicted it.","0\\end{pmatrix}.\n\\]\nThe difference \\(PTQTPx=(0,-1,0)^{\\mathsf T}\\) is the return of the discarded coordinate. Equation~\\eqref{eq:return} predicted it.","For the diagram, write \\(T=\\begin{psmallmatrix}A&B\\\\C&2\\end{psmallmatrix}\\) relative to the retained plane and its complement. Here \\(A=\\begin{psmallmatrix}2&1","C&2\\end{psmallmatrix}\\) relative to the retained plane and its complement. Here \\(A=\\begin{psmallmatrix}2&1\\\\0&3\\end{psmallmatrix}\\), \\(B=(0,1)^{\\mathsf T}\\), and \\(C=(1,-1)\\). The two paths contribute \\(A^2(1,2)^{\\mathsf T}\\) and \\(BC(1,2)^{\\mathsf T}\\) to their sum.","0&3\\end{psmallmatrix}\\), \\(B=(0,1)^{\\mathsf T}\\), and \\(C=(1,-1)\\). The two paths contribute \\(A^2(1,2)^{\\mathsf T}\\) and \\(BC(1,2)^{\\mathsf T}\\) to their sum.","The general packet formulation uses a finite family of pairwise annihilating projectors whose sum is the identity. Write \\(T_{ij}=P_iTP_j\\). Then\n\\[\nT=\\sum_{i,j}T_{ij},\n\\qquad\n(ST)_{ik}=\\sum_j S_{ij}T_{jk}.\n\\]\nComposition follows paths through intermediate packets. The laws are familiar block algebra. Their methodological value is that they provide an exact interface for tracking what a decomposition retains and what it omits. Realization Calculus uses such interfaces to organize more specialized constructions without pretending that an arithmetic lane and an analytic subspace have identical additional properties.\\sourceref{rc}","Invariance supplies a sufficient answer in this setting. If \\(QTP=0\\), then \\(T\\) never sends a retained input into the complementary space, so the retained space is invariant and repeated restriction is exact. The condition is stronger than vanishing on one selected input. It is also stronger than the disappearance of one particular return term. A method that records the operator and its domain can distinguish these statements instead of treating them as interchangeable successes.","The discrepancy can do more than refute a simplification. It can tell us how to construct an exact replacement. Write a general block evolution as\n\\[\n u_{n+1}=Au_n+Bv_n,\\qquad v_{n+1}=Cu_n+Ev_n,\n\\]\nwhere \\(u_n\\) is retained and \\(v_n\\) is complementary. Suppose first that \\(v_0=0\\). Repeated substitution in the second equation gives\n\\[\n v_n=\\sum_{j=0}^{n-1}E^{n-1-j}Cu_j.\n\\]\nFor \\(n=0\\) the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs ev","The discrepancy can do more than refute a simplification. It can tell us how to construct an exact replacement. Write a general block evolution as\n\\[\n u_{n+1}=Au_n+Bv_n,\\qquad v_{n+1}=Cu_n+Ev_n,\n\\]\nwhere \\(u_n\\) is retained and \\(v_n\\) is complementary. Suppose first that \\(v_0=0\\). Repeated substitution in the second equation gives\n\\[\n v_n=\\sum_{j=0}^{n-1}E^{n-1-j}Cu_j.\n\\]\nFor \\(n=0\\) the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs every later step","The discrepancy can do more than refute a simplification. It can tell us how to construct an exact replacement. Write a general block evolution as\n\\[\n u_{n+1}=Au_n+Bv_n,\\qquad v_{n+1}=Cu_n+Ev_n,\n\\]\nwhere \\(u_n\\) is retained and \\(v_n\\) is complementary. Suppose first that \\(v_0=0\\). Repeated substitution in the second equation gives\n\\[\n v_n=\\sum_{j=0}^{n-1}E^{n-1-j}Cu_j.\n\\]\nFor \\(n=0\\) the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs every later step. This is a deduction fr","The discrepancy can do more than refute a simplification. It can tell us how to construct an exact replacement. Write a general block evolution as\n\\[\n u_{n+1}=Au_n+Bv_n,\\qquad v_{n+1}=Cu_n+Ev_n,\n\\]\nwhere \\(u_n\\) is retained and \\(v_n\\) is complementary. Suppose first that \\(v_0=0\\). Repeated substitution in the second equation gives\n\\[\n v_n=\\sum_{j=0}^{n-1}E^{n-1-j}Cu_j.\n\\]\nFor \\(n=0\\) the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs every later step. This is a deduction from the block equations, not a fitted correction","The discrepancy can do more than refute a simplification. It can tell us how to construct an exact replacement. Write a general block evolution as\n\\[\n u_{n+1}=Au_n+Bv_n,\\qquad v_{n+1}=Cu_n+Ev_n,\n\\]\nwhere \\(u_n\\) is retained and \\(v_n\\) is complementary. Suppose first that \\(v_0=0\\). Repeated substitution in the second equation gives\n\\[\n v_n=\\sum_{j=0}^{n-1}E^{n-1-j}Cu_j.\n\\]\nFor \\(n=0\\) the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs every later step. This is a deduction from the block equations, not a fitted correction to a single failed experiment.","can tell us how to construct an exact replacement. Write a general block evolution as\n\\[\n u_{n+1}=Au_n+Bv_n,\\qquad v_{n+1}=Cu_n+Ev_n,\n\\]\nwhere \\(u_n\\) is retained and \\(v_n\\) is complementary. Suppose first that \\(v_0=0\\). Repeated substitution in the second equation gives\n\\[\n v_n=\\sum_{j=0}^{n-1}E^{n-1-j}Cu_j.\n\\]\nFor \\(n=0\\) the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs every later step. This is a deduction from the block equations, not a fitted correction to a single failed experiment.","w to construct an exact replacement. Write a general block evolution as\n\\[\n u_{n+1}=Au_n+Bv_n,\\qquad v_{n+1}=Cu_n+Ev_n,\n\\]\nwhere \\(u_n\\) is retained and \\(v_n\\) is complementary. Suppose first that \\(v_0=0\\). Repeated substitution in the second equation gives\n\\[\n v_n=\\sum_{j=0}^{n-1}E^{n-1-j}Cu_j.\n\\]\nFor \\(n=0\\) the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs every later step. This is a deduction from the block equations, not a fitted correction to a single failed experiment.","u_{n+1}=Au_n+Bv_n,\\qquad v_{n+1}=Cu_n+Ev_n,\n\\]\nwhere \\(u_n\\) is retained and \\(v_n\\) is complementary. Suppose first that \\(v_0=0\\). Repeated substitution in the second equation gives\n\\[\n v_n=\\sum_{j=0}^{n-1}E^{n-1-j}Cu_j.\n\\]\nFor \\(n=0\\) the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs every later step. This is a deduction from the block equations, not a fitted correction to a single failed experiment.","v_{n+1}=Cu_n+Ev_n,\n\\]\nwhere \\(u_n\\) is retained and \\(v_n\\) is complementary. Suppose first that \\(v_0=0\\). Repeated substitution in the second equation gives\n\\[\n v_n=\\sum_{j=0}^{n-1}E^{n-1-j}Cu_j.\n\\]\nFor \\(n=0\\) the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs every later step. This is a deduction from the block equations, not a fitted correction to a single failed experiment.","}^{n-1}E^{n-1-j}Cu_j.\n\\]\nFor \\(n=0\\) the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs every later step. This is a deduction from the block equations, not a fitted correction to a single failed experiment.","the sum is empty. The induction step multiplies the preceding sum by \\(E\\) and adds \\(Cu_n\\). Substituting into the first equation therefore proves, for every \\(n\\geq0\\),\n\\begin{equation}\\label{eq:memory}\n u_{n+1}=Au_n+\\sum_{j=0}^{n-1}BE^{n-1-j}Cu_j.\n\\end{equation}\nThe reduced description is exact, but it now carries memory. In the displayed rational example \\(E=2\\), and the first correction is \\(BCu_0=(0,-1)^{\\mathsf T}\\). Equation~\\eqref{eq:memory} gives the correct second observation and governs every later step. This is a deduction from the block equations, not a fitted correction to a single failed experiment.","The requirement \\(v_0=0\\) is substantive. With a nonzero initial complementary state, the formula for \\(v_n\\) contains the additional term \\(E^nv_0\\), and the retained evolution contains \\(BE^nv_0\\). An unknown \\(v_0\\) cannot simply be erased. Either the problem supplies it, the retained data determine its relevant effect, or the prediction must preserve the remaining ambiguity.","A similar issue arises when comparing two sequences of transformations. Let \\(T:X\\to Y\\) and \\(S:Y\\to Z\\) be linear maps, with comparison maps \\(p:X\\to X_0\\), \\(q:Y\\to Y_0\\), and \\(r:Z\\to Z_0\\). Suppose \\(T_0:X_0\\to Y_0\\) and \\(S_0:Y_0\\to Z_0\\) are the proposed reduced maps. Their defects are\n\\[\n\\Delta_T=qT-T_0p,\n\\qquad\n\\Delta_S=rS-S_0q.\n\\]\nAdding and subtracting \\(S_0qT\\) gives\n\\begin{equation}\\label{eq:composition}\n\\Delta_{ST}=rST-S_0T_0p\n=\\Delta_S T+S_0\\Delta_T.\n\\end{equation}\nOne defect is evaluated after \\(T\\). The other is carried forward by \\(S_0\\). Only then do they have the common source and target needed for addition.","aps, with comparison maps \\(p:X\\to X_0\\), \\(q:Y\\to Y_0\\), and \\(r:Z\\to Z_0\\). Suppose \\(T_0:X_0\\to Y_0\\) and \\(S_0:Y_0\\to Z_0\\) are the proposed reduced maps. Their defects are\n\\[\n\\Delta_T=qT-T_0p,\n\\qquad\n\\Delta_S=rS-S_0q.\n\\]\nAdding and subtracting \\(S_0qT\\) gives\n\\begin{equation}\\label{eq:composition}\n\\Delta_{ST}=rST-S_0T_0p\n=\\Delta_S T+S_0\\Delta_T.\n\\end{equation}\nOne defect is evaluated after \\(T\\). The other is carried forward by \\(S_0\\). Only then do they have the common source and target needed for addition.","), \\(q:Y\\to Y_0\\), and \\(r:Z\\to Z_0\\). Suppose \\(T_0:X_0\\to Y_0\\) and \\(S_0:Y_0\\to Z_0\\) are the proposed reduced maps. Their defects are\n\\[\n\\Delta_T=qT-T_0p,\n\\qquad\n\\Delta_S=rS-S_0q.\n\\]\nAdding and subtracting \\(S_0qT\\) gives\n\\begin{equation}\\label{eq:composition}\n\\Delta_{ST}=rST-S_0T_0p\n=\\Delta_S T+S_0\\Delta_T.\n\\end{equation}\nOne defect is evaluated after \\(T\\). The other is carried forward by \\(S_0\\). Only then do they have the common source and target needed for addition.","The distinction is exact in a linear correction problem. Let \\(D:U\\to W\\) be a linear map over a field, and let \\(r\\in W\\) be a defect. An admissible correction is a vector \\(h\\in U\\) satisfying \\(Dh=-r\\). Such a correction exists precisely when the class of \\(r\\) vanishes in the quotient\n\\[\nW/\\operatorname{im}D.\n\\]\nIndeed, vanishing says \\(r=Dg\\) for some \\(g\\), so \\(h=-g\\) corrects it. Conversely, a correction puts \\(r\\) in the image. The quotient records exactly what the available corrections cannot change.","t in a linear correction problem. Let \\(D:U\\to W\\) be a linear map over a field, and let \\(r\\in W\\) be a defect. An admissible correction is a vector \\(h\\in U\\) satisfying \\(Dh=-r\\). Such a correction exists precisely when the class of \\(r\\) vanishes in the quotient\n\\[\nW/\\operatorname{im}D.\n\\]\nIndeed, vanishing says \\(r=Dg\\) for some \\(g\\), so \\(h=-g\\) corrects it. Conversely, a correction puts \\(r\\) in the image. The quotient records exactly what the available corrections cannot change.","Take \\(U=\\mathbb Q\\), \\(W=\\mathbb Q^2\\), and \\(D(u)=(u,u)\\). Only diagonal corrections are allowed. For \\(r=(1,2)\\), the functional\n\\[\n\\lambda(w_1,w_2)=w_2-w_1\n\\]\nvanishes on every possible correction but satisfies \\(\\lambda(r)=1\\). If \\(Dh=-r\\) held, applying \\(\\lambda\\) would give \\(0=-1\\). No additional search is needed. The functional is a short certificate of impossibility for the stated correction problem.","Now allow \\(D'(u,v)=(u,u+v)\\). The choice \\((-1,-1)\\) supplies the correction \\((-1,-2)\\), and the defect disappears. Nothing contradictory has happened. The admissible correction space has changed. That change may be an intelligent way to solve a broader problem, but it is not a solution of the original problem unless the extra freedom was permitted.","For a linear protected quantity \\(p:U\\to V\\), corrections that leave the quantity unchanged lie in \\(\\ker p\\). The relevant correction image is therefore \\(D(\\ker p)\\), not all of \\(D(U)\\). The obstruction must be calculated relative to that smaller image. This is the elementary form of protected repair developed in the realization framework. It explains why two superficially identical defects may have different answers when their preservation requirements differ.\\sourceref{ks,ppt}","The same care applies to small residuals. On the Hilbert space \\(\\ell^2\\), define \\(D(x)_n=x_n/n\\) for positive integers \\(n\\). The map is bounded and injective. For the unit vector \\(e_n\\), however,\n\\[\n\\|De_n\\|=1/n\\longrightarrow0,\n\\qquad\n\\operatorname{dist}(e_n,\\ker D)=1.\n\\]\nThere is no uniform bound \\(\\operatorname{dist}(x,\\ker D)\\leq C\\|Dx\\|\\) on this space. Taking \\(x=e_n\\) would require \\(1\\leq C/n\\) for every \\(n\\). A tiny residual can therefore coexist with a fixed distance from the exact solution set. A stability estimate needs additional structure, not just a smaller numerical threshold.","to small residuals. On the Hilbert space \\(\\ell^2\\), define \\(D(x)_n=x_n/n\\) for positive integers \\(n\\). The map is bounded and injective. For the unit vector \\(e_n\\), however,\n\\[\n\\|De_n\\|=1/n\\longrightarrow0,\n\\qquad\n\\operatorname{dist}(e_n,\\ker D)=1.\n\\]\nThere is no uniform bound \\(\\operatorname{dist}(x,\\ker D)\\leq C\\|Dx\\|\\) on this space. Taking \\(x=e_n\\) would require \\(1\\leq C/n\\) for every \\(n\\). A tiny residual can therefore coexist with a fixed distance from the exact solution set. A stability estimate needs additional structure, not just a smaller numerical threshold.","A small generating-function calculation illustrates the movement from pattern to explanation. Define the signed Euler numbers by the formal power series identity over \\(\\mathbb Q\\),\n\\[\nF(t)=\\operatorname{sech}t=\\sum_{n\\geq0}E_n\\frac{t^n}{n!}.\n\\]\nThe initial coefficients are \\(1,0,-1,0,5,0,-61,0,1385\\). Multiplication by \\(e^t\\) gives transformed coefficients\n\\[\nb_n=\\sum_{k=0}^n\\binom nk E_k.\n\\]\nTheir initial values are \\(1,1,0,-2,0,16,0,-272,0\\). A finite computation suggests that every positive even index has disappeared. The reason is not hidden in the numerical list:\n\\[\ne^t\\operatorname{sech}t\n=\\frac{\\cosh t+\\sinh t}{\\cosh t}\n=1+\\tanh t.\n\\]\nThe series \\(\\tanh t\\) is odd. Thus \\(b_{2j}=0\\) for every \\(j\\geq1\\). Multiplication by \\(e^{-t}\\) recovers the original coefficients, so the disappearance is a reorganization of information, not the destruction of the sequen","A small generating-function calculation illustrates the movement from pattern to explanation. Define the signed Euler numbers by the formal power series identity over \\(\\mathbb Q\\),\n\\[\nF(t)=\\operatorname{sech}t=\\sum_{n\\geq0}E_n\\frac{t^n}{n!}.\n\\]\nThe initial coefficients are \\(1,0,-1,0,5,0,-61,0,1385\\). Multiplication by \\(e^t\\) gives transformed coefficients\n\\[\nb_n=\\sum_{k=0}^n\\binom nk E_k.\n\\]\nTheir initial values are \\(1,1,0,-2,0,16,0,-272,0\\). A finite computation suggests that every positive even index has disappeared. The reason is not hidden in the numerical list:\n\\[\ne^t\\operatorname{sech}t\n=\\frac{\\cosh t+\\sinh t}{\\cosh t}\n=1+\\tanh t.\n\\]\nThe series \\(\\tanh t\\) is odd. Thus \\(b_{2j}=0\\) for every \\(j\\geq1\\). Multiplication by \\(e^{-t}\\) recovers the original coefficients, so the disappearance is a reorganization of information, not the destruction of the sequence.","on calculation illustrates the movement from pattern to explanation. Define the signed Euler numbers by the formal power series identity over \\(\\mathbb Q\\),\n\\[\nF(t)=\\operatorname{sech}t=\\sum_{n\\geq0}E_n\\frac{t^n}{n!}.\n\\]\nThe initial coefficients are \\(1,0,-1,0,5,0,-61,0,1385\\). Multiplication by \\(e^t\\) gives transformed coefficients\n\\[\nb_n=\\sum_{k=0}^n\\binom nk E_k.\n\\]\nTheir initial values are \\(1,1,0,-2,0,16,0,-272,0\\). A finite computation suggests that every positive even index has disappeared. The reason is not hidden in the numerical list:\n\\[\ne^t\\operatorname{sech}t\n=\\frac{\\cosh t+\\sinh t}{\\cosh t}\n=1+\\tanh t.\n\\]\nThe series \\(\\tanh t\\) is odd. Thus \\(b_{2j}=0\\) for every \\(j\\geq1\\). Multiplication by \\(e^{-t}\\) recovers the original coefficients, so the disappearance is a reorganization of information, not the destruction of the sequence.","),\n\\[\nF(t)=\\operatorname{sech}t=\\sum_{n\\geq0}E_n\\frac{t^n}{n!}.\n\\]\nThe initial coefficients are \\(1,0,-1,0,5,0,-61,0,1385\\). Multiplication by \\(e^t\\) gives transformed coefficients\n\\[\nb_n=\\sum_{k=0}^n\\binom nk E_k.\n\\]\nTheir initial values are \\(1,1,0,-2,0,16,0,-272,0\\). A finite computation suggests that every positive even index has disappeared. The reason is not hidden in the numerical list:\n\\[\ne^t\\operatorname{sech}t\n=\\frac{\\cosh t+\\sinh t}{\\cosh t}\n=1+\\tanh t.\n\\]\nThe series \\(\\tanh t\\) is odd. Thus \\(b_{2j}=0\\) for every \\(j\\geq1\\). Multiplication by \\(e^{-t}\\) recovers the original coefficients, so the disappearance is a reorganization of information, not the destruction of the sequence.","ients are \\(1,0,-1,0,5,0,-61,0,1385\\). Multiplication by \\(e^t\\) gives transformed coefficients\n\\[\nb_n=\\sum_{k=0}^n\\binom nk E_k.\n\\]\nTheir initial values are \\(1,1,0,-2,0,16,0,-272,0\\). A finite computation suggests that every positive even index has disappeared. The reason is not hidden in the numerical list:\n\\[\ne^t\\operatorname{sech}t\n=\\frac{\\cosh t+\\sinh t}{\\cosh t}\n=1+\\tanh t.\n\\]\nThe series \\(\\tanh t\\) is odd. Thus \\(b_{2j}=0\\) for every \\(j\\geq1\\). Multiplication by \\(e^{-t}\\) recovers the original coefficients, so the disappearance is a reorganization of information, not the destruction of the sequence.","-61,0,1385\\). Multiplication by \\(e^t\\) gives transformed coefficients\n\\[\nb_n=\\sum_{k=0}^n\\binom nk E_k.\n\\]\nTheir initial values are \\(1,1,0,-2,0,16,0,-272,0\\). A finite computation suggests that every positive even index has disappeared. The reason is not hidden in the numerical list:\n\\[\ne^t\\operatorname{sech}t\n=\\frac{\\cosh t+\\sinh t}{\\cosh t}\n=1+\\tanh t.\n\\]\nThe series \\(\\tanh t\\) is odd. Thus \\(b_{2j}=0\\) for every \\(j\\geq1\\). Multiplication by \\(e^{-t}\\) recovers the original coefficients, so the disappearance is a reorganization of information, not the destruction of the sequence.","ation by \\(e^t\\) gives transformed coefficients\n\\[\nb_n=\\sum_{k=0}^n\\binom nk E_k.\n\\]\nTheir initial values are \\(1,1,0,-2,0,16,0,-272,0\\). A finite computation suggests that every positive even index has disappeared. The reason is not hidden in the numerical list:\n\\[\ne^t\\operatorname{sech}t\n=\\frac{\\cosh t+\\sinh t}{\\cosh t}\n=1+\\tanh t.\n\\]\nThe series \\(\\tanh t\\) is odd. Thus \\(b_{2j}=0\\) for every \\(j\\geq1\\). Multiplication by \\(e^{-t}\\) recovers the original coefficients, so the disappearance is a reorganization of information, not the destruction of the sequence.","rmed coefficients\n\\[\nb_n=\\sum_{k=0}^n\\binom nk E_k.\n\\]\nTheir initial values are \\(1,1,0,-2,0,16,0,-272,0\\). A finite computation suggests that every positive even index has disappeared. The reason is not hidden in the numerical list:\n\\[\ne^t\\operatorname{sech}t\n=\\frac{\\cosh t+\\sinh t}{\\cosh t}\n=1+\\tanh t.\n\\]\nThe series \\(\\tanh t\\) is odd. Thus \\(b_{2j}=0\\) for every \\(j\\geq1\\). Multiplication by \\(e^{-t}\\) recovers the original coefficients, so the disappearance is a reorganization of information, not the destruction of the sequence.","Residue selectors make the mechanism more general. Let \\(F(t)=\\sum_{n\\geq0}a_nt^n\\) be a formal series over \\(\\mathbb C\\), let \\(\\zeta\\) be a primitive \\(m\\)-th root of unity, and choose \\(0\\leq r<m\\). Then\n\\[\n\\frac1m\\sum_{j=0}^{m-1}\\zeta^{-rj}F(\\zeta^jt)\n=\\sum_{n\\equiv r\\;({\\rm mod}\\,m)}a_nt^n.\n\\]\nFor the coefficient of \\(t^n\\), the multiplier is the average of \\(\\zeta^{j(n-r)}\\). It is one when \\(m\\) divides \\(n-r\\), and zero otherwise by the finite geometric-sum identity. The filter selects a lane of coefficients exactly. Because the argument is coefficientwise, it needs no analytic convergence assumption.","sidue selectors make the mechanism more general. Let \\(F(t)=\\sum_{n\\geq0}a_nt^n\\) be a formal series over \\(\\mathbb C\\), let \\(\\zeta\\) be a primitive \\(m\\)-th root of unity, and choose \\(0\\leq r<m\\). Then\n\\[\n\\frac1m\\sum_{j=0}^{m-1}\\zeta^{-rj}F(\\zeta^jt)\n=\\sum_{n\\equiv r\\;({\\rm mod}\\,m)}a_nt^n.\n\\]\nFor the coefficient of \\(t^n\\), the multiplier is the average of \\(\\zeta^{j(n-r)}\\). It is one when \\(m\\) divides \\(n-r\\), and zero otherwise by the finite geometric-sum identity. The filter selects a lane of coefficients exactly. Because the argument is coefficientwise, it needs no analytic convergence assumption.","s make the mechanism more general. Let \\(F(t)=\\sum_{n\\geq0}a_nt^n\\) be a formal series over \\(\\mathbb C\\), let \\(\\zeta\\) be a primitive \\(m\\)-th root of unity, and choose \\(0\\leq r<m\\). Then\n\\[\n\\frac1m\\sum_{j=0}^{m-1}\\zeta^{-rj}F(\\zeta^jt)\n=\\sum_{n\\equiv r\\;({\\rm mod}\\,m)}a_nt^n.\n\\]\nFor the coefficient of \\(t^n\\), the multiplier is the average of \\(\\zeta^{j(n-r)}\\). It is one when \\(m\\) divides \\(n-r\\), and zero otherwise by the finite geometric-sum identity. The filter selects a lane of coefficients exactly. Because the argument is coefficientwise, it needs no analytic convergence assumption.","This is also where changing the coefficient setting matters. The displayed average uses roots of unity and division by \\(m\\). Transporting it into an integral or \\(p\\)-adic argument requires control of those operations, especially when \\(p\\) divides \\(m\\). A valid identity over \\(\\mathbb C\\) does not by itself supply the integral estimate one hopes to extract from it.","Axioms require a separate audit. Lean permits new axioms and records proof dependencies. Postulating \\(\\mathrm{False}\\) allows any proposition to be derived. The derivation is valid relative to that assumption. A project that disallows the assumption must reject the dependency.\\sourceref{leanaxioms}","Ordinary hypotheses are different from global axioms. The implication \\(\\mathrm{False}\\Rightarrow P\\) needs no newly postulated axiom, yet cannot establish \\(P\\) without the impossible premise. Less conspicuous hypotheses can conceal the same application gap. The review must therefore inspect the theorem's quantifiers and show that the required data are available where the result is used. A legitimate conditional theorem need not supply every application itself. Its conditional scope must remain visible.","The finite model uses justification hypergraphs because ordinary arrows can hide an important distinction. A step may require both \\(A\\) and \\(B\\) to obtain \\(L\\). Another conclusion may have two independent routes, either of which is sufficient. The former is a conjunction of premises. The latter is an alternative between justifications. A faithful record must preserve which kind of dependency is present.","For example, take certified steps \\(\\{A,B\\}\\Rightarrow L\\) and \\(L\\Rightarrow T\\). They form one certificate for the target \\(T\\). A second route might use \\(A\\Rightarrow M\\) and \\(\\{M,B\\}\\Rightarrow T\\). A candidate extension from \\(M\\) to a new target \\(U\\) can remain visible without becoming part of either certificate. It may be valuable precisely because it suggests what to investigate next, but its value does not make it valid.","For the running example, the durable record is short but specific. The task is to reproduce retained observations of the full evolution from the declared initial state. The failed candidate is iteration of \\(A\\) alone. The counterexample occurs at the second observation. The replacement is the memory law, with its initial-state hypothesis and induction proof. A later researcher can now change the question intelligently: ask for a shorter memory, a bound on a truncated history, or a sufficient statistic that can be updated cheaply. Keeping only the final matrix product would conceal this direction of inquiry.","from factorization. The return, memory, and defect identities are linear statements. The memory formula retains the declared initial complementary state, including its extra contribution when that state is nonzero. The correction alternative is relative to the specified correction image. The finite-prefix example uses inclusion maps, and the stability example uses the stated bounded operator on \\(\\ell^2\\). The roots-of-unity identity is coefficientwise over \\(\\mathbb C[[t]]\\). Keeping these settings explicit is part of the argument.","e linear statements. The memory formula retains the declared initial complementary state, including its extra contribution when that state is nonzero. The correction alternative is relative to the specified correction image. The finite-prefix example uses inclusion maps, and the stability example uses the stated bounded operator on \\(\\ell^2\\). The roots-of-unity identity is coefficientwise over \\(\\mathbb C[[t]]\\). Keeping these settings explicit is part of the argument."]}