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Proof step 3 from proof abs-avols-size (previous step)

# vols[v] < # avols[v], m < # avols[v], v ∈ avols, # vols[v] ≤ m, ∀ 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))) ⊦

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: v

Additional information

Used simplifier rules: lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m +1 = m + 1; (s)% ⊦ ¬ n < n; (s) Used terms: v