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Proof step 3 from proof abs-vols-unmapped (previous step)

vols[v][m] = unmapped, avols[v][m].mapped?, m < # vols[v], v ∈ vols, ∀ 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 obtained the following new goals

Proof step 4:
v ∈ vols ↔ v ∈ avols, vols[v][m] = unmapped, avols[v][m].mapped?, m < # vols[v], v ∈ vols, 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)), ∀ 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))) ⊦

Rule argument

pos: left 5 terms: v

Additional information

Used terms: v