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Proof step 7 from proof abs-erase-emptyq (previous step)

v0 ∈ vols ↔ v0 ∈ avols, apebs[vols[v, vols[v][m, unmapped]][v0][m0].pnum] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m < # vols[v], v ∈ vols, v0 ∈ vols, v0 ∈ avols → # vols[v0] = # avols[v0] ∧ (∀ m. m < # vols[v0] → (vols[v0][m] = unmapped → ¬ avols[v0][m].mapped?) ∧ ( vols[v0][m] ≠ unmapped → avols[v0][m].mapped? ∧ avols[v0][m].data = apebs[vols[v0][m].pnum].data ∧ apebs[vols[v0][m].pnum].fill ≤ avols[v0][m].fill) ∧ (avols[v0][m] = erased → ¬ v0 × m ∈ q)) ⊦ (v0 ∈ vols[v, vols[v][m, unmapped]] ↔ v0 ∈ avols[v, avols[v][m, erased]]) ∧ ( v0 ∈ vols[v, vols[v][m, unmapped]] → # vols[v, vols[v][m, unmapped]][v0] = # avols[v, avols[v][m, erased]][v0] ∧ ( m0 < # vols[v, vols[v][m, unmapped]][v0] → (vols[v, vols[v][m, unmapped]][v0][m0] = unmapped → ¬ avols[v, avols[v][m, erased]][v0][m0].mapped?) ∧ ( vols[v, vols[v][m, unmapped]][v0][m0] ≠ unmapped → avols[v, avols[v][m, erased]][v0][m0].mapped? ∧ avols[v, avols[v][m, erased]][v0][m0].data = buf ∧ m1 ≤ avols[v, avols[v][m, erased]][v0][m0].fill)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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