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Proof step 8 from proof abs-queue-notin (previous step)

# vols[v0] = # avols[v0], apebs[vols[v0][m0].pnum] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # avols[v0], v0 ∈ avols, v0 ∈ vols, ∀ m0. m0 < # avols[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) '), ∀ 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, (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 closed the goal!

Rule argument

pos: left 6 terms: m0

Additional information

Used simplifier rules: ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).data = buf; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).fill = m; (s)% ⊦ lref ∈ q + eq-entry(n, lref0) ' ↔ lref ∈ q ∨ lref = lref0; (ss)%a: ⊦ (q + q0) + q1 = q + q0 + q1; from specification list[erasequeue-base] Used terms: m0