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

# vols[v] = # avols[v], apebs[vols[v][m].pnum] = mkpeb(aehdr, avhdr, buf, m0, boolvar), m < # avols[v], v ∈ avols, v ∈ vols, q1 ⊆ q0, ∀ m. m < # avols[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 ∈ q0), ∀ 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 ∈ q0))) ⊦ (vols[v][m] = unmapped → ¬ avols[v][m].mapped?) ∧ (vols[v][m] ≠ unmapped → avols[v][m].mapped? ∧ avols[v][m].data = buf ∧ m0 ≤ avols[v][m].fill) ∧ (avols[v][m] = erased → ¬ v × m ∈ q1)

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: m

Additional information

Used simplifier rules: ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).data = buf; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).fill = m; (s)% ⊦ q1 ⊆ q0 ∧ lref ∈ q1 → lref ∈ q0; (ss) Used terms: m