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Proof step 6 from proof abs-queue-notin (previous step)

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

Heuristic quantifier applied rule all left

And obtained the following new goals

Rule argument

pos: left 2 terms: v0

Additional information

Used terms: v0