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Proof step 20 from proof vols-le-antisym (previous step)

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

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n

Additional information

Used terms: n