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Proof step 23 from proof vols-le-antisym (previous step)

v ∈ vols0, ∀ v. (v ∈ vols0 ↔ v ∈ vols1) ∧ (v ∈ vols0 → # vols0[v] = # vols1[v] ∧ (∀ m. m < # vols0[v] ∧ vols0[v][m] ≠ unmapped → vols0[v][m] = vols1[v][m])), ∀ v. (v ∈ vols1 ↔ v ∈ vols0) ∧ (v ∈ vols1 → # vols1[v] = # vols0[v] ∧ (∀ m. m < # vols1[v] ∧ vols1[v][m] ≠ unmapped → vols1[v][m] = vols0[v][m])) ⊦ v ∈ vols1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: v

Additional information

Used terms: v