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Proof step 2 from proof abs-unmap-twice (previous step)

vols[v][m] = unmapped, m < # vols[v], abs(avols, apebs, vols, q), v ∈ vols ⊦ abs(avols[v, avols[v][m, unmapped]], apebs, vols, q)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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))));% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic