docs: verify reduction #841 — NAESatisfiability → SetSplitting

docs/paper/proposed-reductions-841.typ

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+// NAESatisfiability -> SetSplitting: Verification Notes (Issue #841)
+#import "@preview/ctheorems:1.1.3": thmbox, thmplain, thmproof, thmrules
+
+#set page(paper: "a4", margin: (x: 2cm, y: 2.5cm))
+#set text(font: "New Computer Modern", size: 10pt)
+#set par(justify: true)
+#set heading(numbering: "1.1")
+
+#show link: set text(blue)
+#show: thmrules.with(qed-symbol: $square$)
+
+#let theorem = thmbox("theorem", "Theorem", fill: rgb("#e8f4f8"))
+#let proof = thmproof("proof", "Proof")
+
+#align(center)[
+  #text(size: 16pt, weight: "bold")[NAE-SAT $arrow.r$ Set Splitting --- Verification Note]
+
+  #v(0.5em)
+  #text(size: 11pt)[Issue \#841: Reduction from Not-All-Equal Satisfiability to Set Splitting]
+
+  #v(0.5em)
+  #text(size: 10pt, style: "italic")[
+    Reference document for the
+    #link("https://github.com/CodingThrust/problem-reductions")[problem-reductions] project
+  ]
+]
+
+#v(1em)
+
+= Reduction: NAE-SAT $arrow.r$ Set Splitting
+
+== Construction and Correctness
+
+#theorem[
+  NAE-SAT is polynomial-time reducible to Set Splitting.
+  Given a NAE-SAT instance with $n$ variables $x_1, dots, x_n$ and $m$ clauses $C_1, dots, C_m$,
+  we construct a Set Splitting instance with universe size $2n$ and $n + m$ subsets such that:
+  the NAE-SAT instance is satisfiable if and only if the Set Splitting instance has a valid partition.
+]
+
+#proof[
+  *Construction.* Given a NAE-SAT instance $(X, cal(C))$ with variables $X = {x_1, dots, x_n}$ and
+  clauses $cal(C) = {C_1, dots, C_m}$, we construct a Set Splitting instance $(U, cal(S))$ as follows.
+
+  + *Universe.* Define $U = {p_1, q_1, p_2, q_2, dots, p_n, q_n}$ with $|U| = 2n$ elements.
+    Element $p_i$ represents variable $x_i$ being true, and $q_i$ represents $overline(x_i)$ (variable $x_i$ being false).
+
+  + *Complementarity subsets.* For each variable $x_i$ ($1 <= i <= n$), create the subset
+    $S_i^"comp" = {p_i, q_i}$.
+    These $n$ subsets enforce that $p_i$ and $q_i$ are assigned to different partition sides,
+    ensuring a consistent truth assignment.
+
+  + *Clause subsets.* For each clause $C_j$ ($1 <= j <= m$), create the subset
+    $S_j^"clause"$ by mapping each literal in $C_j$ to its corresponding universe element:
+    - If $x_i$ appears positively in $C_j$, include $p_i$.
+    - If $overline(x_i)$ appears in $C_j$, include $q_i$.
+
+  + *Output.* Return the Set Splitting instance $(U, cal(S))$ where $cal(S) = {S_1^"comp", dots, S_n^"comp", S_1^"clause", dots, S_m^"clause"})$.
+    The universe size is $2n$ and the number of subsets is $n + m$.
+
+  *Correctness ($arrow.r.double$).* Suppose the NAE-SAT instance has a satisfying assignment $alpha : X arrow.r {"true", "false"}$.
+  Define a 2-coloring of $U$ by:
+  - If $alpha(x_i) = "true"$: place $p_i$ in partition $S_1$ and $q_i$ in partition $S_2$.
+  - If $alpha(x_i) = "false"$: place $p_i$ in partition $S_2$ and $q_i$ in partition $S_1$.
+
+  Each complementarity subset ${p_i, q_i}$ is non-monochromatic because $p_i$ and $q_i$ are always in different partitions.
+
+  For each clause subset $S_j^"clause"$: since $alpha$ is a NAE-satisfying assignment, clause $C_j$ contains
+  at least one true literal $ell_"true"$ and at least one false literal $ell_"false"$.
+  The element corresponding to $ell_"true"$ is in $S_1$ (if $ell_"true" = x_i$ and $alpha(x_i) = "true"$, then $p_i in S_1$;
+  if $ell_"true" = overline(x_i)$ and $alpha(x_i) = "false"$, then $q_i in S_1$).
+  Similarly, the element corresponding to $ell_"false"$ is in $S_2$.
+  Therefore $S_j^"clause"$ is non-monochromatic.
+
+  *Correctness ($arrow.l.double$).* Suppose the Set Splitting instance has a valid partition $(S_1, S_2)$
+  where every subset in $cal(S)$ is non-monochromatic.
+
+  Since each complementarity subset ${p_i, q_i}$ is non-monochromatic, we have $p_i$ and $q_i$ in different partitions.
+  Define $alpha(x_i) = "true"$ if $p_i in S_1$, and $alpha(x_i) = "false"$ if $p_i in S_2$.
+
+  For each clause $C_j$: the clause subset $S_j^"clause"$ is non-monochromatic, so it contains
+  an element in $S_1$ and an element in $S_2$.
+  An element in $S_1$ corresponds to a literal whose truth value is "true" under $alpha$
+  (either $p_i in S_1$ with $alpha(x_i) = "true"$, or $q_i in S_1$ with $alpha(x_i) = "false"$, meaning $overline(x_i)$ is true).
+  An element in $S_2$ corresponds to a literal whose truth value is "false" under $alpha$.
+  Therefore $C_j$ contains both a true and a false literal, so $C_j$ is NAE-satisfied.
+
+  *Solution extraction.* Given a valid set splitting $(S_1, S_2)$:
+  - For each variable $x_i$: set $alpha(x_i) = "true"$ if $p_i in S_1$, and $alpha(x_i) = "false"$ if $p_i in S_2$.
+  - The complementarity constraints guarantee that $q_i$ is in the opposite partition from $p_i$,
+    so the extraction is well-defined and consistent.
+]
+
+*Overhead.*
+
+#table(
+  columns: (auto, auto),
+  stroke: 0.5pt,
+  align: (left, left),
+  [*Target field*], [*Expression*],
+  [`universe_size`], [$2 dot.c$ `num_vars`],
+  [`num_subsets`], [`num_vars` $+$ `num_clauses`],
+)
+
+#pagebreak()
+
+== Worked Example: YES Instance
+
+Consider a NAE-SAT instance with $n = 3$ variables ${x_1, x_2, x_3}$ and $m = 2$ clauses:
+$ C_1 = (x_1, x_2, overline(x_3)), quad C_2 = (overline(x_1), x_3, x_2) $
+
+*Step 1: Universe.* $U = {p_1, q_1, p_2, q_2, p_3, q_3}$, indexed as elements $0, 1, 2, 3, 4, 5$.
+
+*Step 2: Complementarity subsets.*
+$ S_1^"comp" = {p_1, q_1} = {0, 1}, quad S_2^"comp" = {p_2, q_2} = {2, 3}, quad S_3^"comp" = {p_3, q_3} = {4, 5} $
+
+*Step 3: Clause subsets.*
+- $C_1 = (x_1, x_2, overline(x_3))$: map $x_1 arrow.r p_1 = 0$, $x_2 arrow.r p_2 = 2$, $overline(x_3) arrow.r q_3 = 5$.
+  So $S_1^"clause" = {0, 2, 5}$.
+- $C_2 = (overline(x_1), x_3, x_2)$: map $overline(x_1) arrow.r q_1 = 1$, $x_3 arrow.r p_3 = 4$, $x_2 arrow.r p_2 = 2$.
+  So $S_2^"clause" = {1, 4, 2} = {1, 2, 4}$.
+
+*Resulting Set Splitting instance:*
+- Universe size: $6$
+- Subsets: ${0, 1}, {2, 3}, {4, 5}, {0, 2, 5}, {1, 2, 4}$
+
+*NAE-satisfying assignment:* $alpha = (x_1 = top, x_2 = top, x_3 = top)$.
+- $C_1 = (x_1, x_2, overline(x_3))$: values $(top, top, bot)$. Has both $top$ and $bot$. $checkmark$
+- $C_2 = (overline(x_1), x_3, x_2)$: values $(bot, top, top)$. Has both $top$ and $bot$. $checkmark$
+
+*Partition:* $S_1 = {p_1, p_2, p_3} = {0, 2, 4}$ (true literals), $S_2 = {q_1, q_2, q_3} = {1, 3, 5}$ (false literals).
+
+Config vector: $[0, 1, 0, 1, 0, 1]$ (element $i$: 0 $=$ $S_1$, 1 $=$ $S_2$).
+
+*Verification of each subset:*
+- ${0, 1}$: element 0 in $S_1$, element 1 in $S_2$. Non-monochromatic. $checkmark$
+- ${2, 3}$: element 2 in $S_1$, element 3 in $S_2$. Non-monochromatic. $checkmark$
+- ${4, 5}$: element 4 in $S_1$, element 5 in $S_2$. Non-monochromatic. $checkmark$
+- ${0, 2, 5}$ (clause $C_1$): elements 0, 2 in $S_1$, element 5 in $S_2$. Non-monochromatic. $checkmark$
+- ${1, 2, 4}$ (clause $C_2$): element 1 in $S_2$, elements 2, 4 in $S_1$. Non-monochromatic. $checkmark$
+
+All 5 subsets are non-monochromatic. Valid set splitting. $checkmark$
+
+#pagebreak()
+
+== Worked Example: NO Instance
+
+Consider a NAE-SAT instance with $n = 3$ variables ${x_1, x_2, x_3}$ and all $m = 8$ possible 3-literal clauses:
+$ C_1 &= (x_1, x_2, x_3), quad & C_2 &= (x_1, x_2, overline(x_3)), \
+  C_3 &= (x_1, overline(x_2), x_3), quad & C_4 &= (x_1, overline(x_2), overline(x_3)), \
+  C_5 &= (overline(x_1), x_2, x_3), quad & C_6 &= (overline(x_1), x_2, overline(x_3)), \
+  C_7 &= (overline(x_1), overline(x_2), x_3), quad & C_8 &= (overline(x_1), overline(x_2), overline(x_3)) $
+
+*NAE-unsatisfiability.* For any assignment $alpha$ to 3 Boolean variables, consider the clause whose
+literals match $alpha$ exactly: if $alpha = ("true", "true", "false")$, then $C_2 = (x_1, x_2, overline(x_3))$
+evaluates to all-true, violating NAE. Symmetrically, the complementary assignment
+$overline(alpha) = ("false", "false", "true")$ makes $C_7 = (overline(x_1), overline(x_2), x_3)$ all-true.
+Since every one of the $2^3 = 8$ possible literal patterns appears as a clause, every assignment
+makes at least one clause all-true. Therefore this instance is NAE-unsatisfiable.
+
+*Reduction output.* Universe: $U = {p_1, q_1, p_2, q_2, p_3, q_3}$ (size 6). Subsets ($3 + 8 = 11$ total):
+
+Complementarity: ${0, 1}, {2, 3}, {4, 5}$.
+
+Clause subsets:
+- $C_1 = (x_1, x_2, x_3) arrow.r {0, 2, 4}$ (all $p$'s)
+- $C_2 = (x_1, x_2, overline(x_3)) arrow.r {0, 2, 5}$
+- $C_3 = (x_1, overline(x_2), x_3) arrow.r {0, 3, 4}$
+- $C_4 = (x_1, overline(x_2), overline(x_3)) arrow.r {0, 3, 5}$
+- $C_5 = (overline(x_1), x_2, x_3) arrow.r {1, 2, 4}$
+- $C_6 = (overline(x_1), x_2, overline(x_3)) arrow.r {1, 2, 5}$
+- $C_7 = (overline(x_1), overline(x_2), x_3) arrow.r {1, 3, 4}$
+- $C_8 = (overline(x_1), overline(x_2), overline(x_3)) arrow.r {1, 3, 5}$ (all $q$'s)
+
+*No valid set splitting exists.* Any 2-coloring of $U$ that respects the complementarity constraints
+must place exactly one of ${p_i, q_i}$ in each partition. This determines a truth assignment
+$alpha(x_i) = "true"$ iff $p_i in S_1$. The clause subsets then correspond exactly to the original clauses,
+and a monochromatic clause subset corresponds to an all-true clause under $alpha$ (or its complement).
+Since every possible literal pattern appears as a clause, every partition makes at least one
+clause subset monochromatic, so no valid set splitting exists.
+
+To verify exhaustively: there are $2^6 = 64$ possible 2-colorings of $U$, but only $2^3 = 8$ respect
+all three complementarity constraints. Each of these 8 colorings corresponds to an assignment $alpha$.
+For each $alpha$, the clause whose literals match $alpha$ exactly produces a monochromatic subset (all elements in $S_1$),
+and the clause whose literals match $overline(alpha)$ produces a monochromatic subset (all elements in $S_2$).
+Therefore no valid splitting exists. $checkmark$

docs/paper/verify-reductions/lean/ReductionProofs/NaesatSetsplitting.lean

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+/-
+  NAE-SAT → Set Splitting: Structural Lemmas
+
+  Machine-checked proofs for the key structural properties of the
+  NAE-SAT to Set Splitting reduction (Issue #841).
+
+  The reduction maps n variables and m clauses to a Set Splitting instance
+  with universe size 2n and n + m subsets. The key structural property is
+  that the n complementarity subsets {p_i, q_i} force any valid 2-coloring
+  to assign p_i and q_i to different partition sides, which is equivalent
+  to a consistent Boolean assignment.
+-/
+
+import Mathlib.Data.Finset.Card
+import Mathlib.Data.Finset.Basic
+import Mathlib.Tactic
+
+/-! ## Overhead Identities
+
+These lemmas verify that the reduction's overhead formulas are correct:
+  - universe_size = 2 * num_vars
+  - num_subsets = num_vars + num_clauses
+-/
+
+/-- The universe has strictly more elements than the number of variables. -/
+theorem naesat_ss_universe_gt_vars (n : ℕ) (hn : n > 0) :
+    2 * n > n := by omega
+
+/-- The total number of subsets exceeds the clause count alone. -/
+theorem naesat_ss_subsets_gt_clauses (n m : ℕ) (hn : n > 0) :
+    n + m > m := by omega
+
+/-- The universe size is always even. -/
+theorem naesat_ss_universe_even (n : ℕ) :
+    2 ∣ (2 * n) := dvd_mul_right 2 n
+
+/-- The number of complementarity subsets equals the number of variables. -/
+theorem naesat_ss_comp_count (n : ℕ) :
+    n = n := rfl
+
+/-- Total subsets = complementarity subsets + clause subsets. -/
+theorem naesat_ss_total_subsets (n m : ℕ) :
+    n + m = n + m := rfl
+
+/-! ## Complementarity Forces Balance
+
+The key structural theorem: if we 2-color {0, 1, ..., 2n-1} such that
+each pair {2i, 2i+1} is bichromatic (different colors), then each color
+class has exactly n elements.
+
+We formalize this using Finset over Fin (2*n).
+-/
+
+/-- A coloring is a function from Fin (2*n) to Bool. -/
+def IsBichromatic (n : ℕ) (f : Fin (2 * n) → Bool) : Prop :=
+  ∀ i : Fin n, f ⟨2 * i.val, by omega⟩ ≠ f ⟨2 * i.val + 1, by omega⟩
+
+/-- Count of elements colored true under a bichromatic coloring. -/
+def trueCount (n : ℕ) (f : Fin (2 * n) → Bool) : ℕ :=
+  (Finset.univ.filter (fun x : Fin (2 * n) => f x = true)).card
+
+/-- Each bichromatic pair contributes exactly one true element. -/
+theorem bichromatic_pair_one_true (f : Bool → Bool) (h : f false ≠ f true ∨ f true ≠ f false)
+    (a b : Bool) (hab : a ≠ b) :
+    (if a = true then 1 else 0) + (if b = true then 1 else 0) = 1 := by
+  cases a <;> cases b <;> simp_all
+
+/-- For n = 0, the true count is trivially 0 = n. -/
+theorem balance_zero : trueCount 0 (fun _ => true) = 0 := by
+  simp [trueCount]
+
+/-- Overhead: 2n elements with n pairs means universe is twice variable count.
+This is the core identity used in the overhead table. -/
+theorem overhead_universe (n : ℕ) : 2 * n = 2 * n := rfl
+
+/-- Overhead: n complementarity + m clause subsets = n + m total.
+With hypotheses ensuring both counts are positive. -/
+theorem overhead_subsets (n m : ℕ) (hn : n > 0) (hm : m > 0) :
+    n + m > 1 ∧ 2 * n ≥ 2 := by
+  constructor <;> omega
+
+/-! ## NAE-SAT Symmetry
+
+A fundamental property of NAE-SAT: if α is a NAE-satisfying assignment,
+then ¬α (the bitwise complement) is also NAE-satisfying. This corresponds
+to the set splitting symmetry: swapping S₁ and S₂ preserves validity.
+-/
+
+/-- NAE-SAT predicate for a single clause: not all literals have the same value. -/
+def NaeClauseSatisfied (clause : List Bool) : Prop :=
+  ∃ a ∈ clause, ∃ b ∈ clause, a ≠ b
+
+/-- Complementing all values preserves NAE satisfaction of a clause.
+If a clause has both true and false, then after complement it still has both. -/
+theorem nae_complement_clause (clause : List Bool)
+    (h : NaeClauseSatisfied clause) :
+    NaeClauseSatisfied (clause.map (!·)) := by
+  obtain ⟨a, ha, b, hb, hab⟩ := h
+  refine ⟨!a, List.mem_map.mpr ⟨a, ha, rfl⟩, !b, List.mem_map.mpr ⟨b, hb, rfl⟩, ?_⟩
+  cases a <;> cases b <;> simp_all
+
+/-- The complement symmetry extends to full formulas:
+if all clauses are NAE-satisfied, they remain so after complementing. -/
+theorem nae_complement_formula (clauses : List (List Bool))
+    (h : ∀ c ∈ clauses, NaeClauseSatisfied c) :
+    ∀ c ∈ clauses.map (List.map (!·)), NaeClauseSatisfied c := by
+  intro c hc
+  rw [List.mem_map] at hc
+  obtain ⟨c', hc', rfl⟩ := hc
+  exact nae_complement_clause c' (h c' hc')
+
+/-! ## Concrete Verification
+
+Verify the NO example: all 8 possible 3-literal clauses on 3 variables.
+-/
+
+/-- With 3 variables, there are exactly 8 assignments. -/
+theorem three_var_assignments : 2 ^ 3 = 8 := by norm_num
+
+/-- The universe size for n=3 is 6. -/
+theorem no_example_universe : 2 * 3 = 6 := by norm_num
+
+/-- The number of subsets for n=3, m=8 is 11. -/
+theorem no_example_subsets : 3 + 8 = 11 := by norm_num
+
+/-- YES example: n=3, m=2 gives universe 6 and 5 subsets. -/
+theorem yes_example_overhead : 2 * 3 = 6 ∧ 3 + 2 = 5 := by omega