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Proof step 2 from proof abs-queue-notin (previous step)

⊦ abs(avols, apebs, vols, q + eq-entry(n, v × m) ') ↔ abs(avols, apebs, vols, q), v ∈ vols

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ (∀ v0. (v0 ∈ vols ↔ v0 ∈ avols) ∧ ( v0 ∈ vols → # vols[v0] = # avols[v0] ∧ (∀ m0. m0 < # vols[v0] → (vols[v0][m0] = unmapped → ¬ avols[v0][m0].mapped?) ∧ ( vols[v0][m0] ≠ unmapped → avols[v0][m0].mapped? ∧ avols[v0][m0].data = apebs[vols[v0][m0].pnum].data ∧ apebs[vols[v0][m0].pnum].fill ≤ avols[v0][m0].fill) ∧ (avols[v0][m0] = erased → ¬ v0 × m0 ∈ q + eq-entry(n, v × m) ')))) ↔ (∀ v. (v ∈ vols ↔ v ∈ avols) ∧ ( v ∈ vols → # vols[v] = # avols[v] ∧ (∀ m. m < # vols[v] → (vols[v][m] = unmapped → ¬ avols[v][m].mapped?) ∧ ( vols[v][m] ≠ unmapped → avols[v][m].mapped? ∧ avols[v][m].data = apebs[vols[v][m].pnum].data ∧ apebs[vols[v][m].pnum].fill ≤ avols[v][m].fill) ∧ (avols[v][m] = erased → ¬ v × m ∈ q)))), v ∈ vols

Rule argument

Additional information

Used simplifier rules: abs-def: ⊦ abs(avols, apebs, vols, q) ↔ (∀ v. (v ∈ vols ↔ v ∈ avols) ∧ ( v ∈ vols → # vols[v] = # avols[v] ∧ (∀ m. m < # vols[v] → (vols[v][m] = unmapped → ¬ avols[v][m].mapped?) ∧ ( vols[v][m] ≠ unmapped → avols[v][m].mapped? ∧ avols[v][m].data = apebs[vols[v][m].pnum].data ∧ apebs[vols[v][m].pnum].fill ≤ avols[v][m].fill) ∧ (avols[v][m] = erased → ¬ v × m ∈ q))));