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Proof step 9 from proof vols-le-trans (previous step)

# vols1[v] = # vols2[v], # vols0[v] = # vols2[v], m < # vols2[v], v ∈ vols2, v ∈ vols1, v ∈ vols0, vols1[v][m] ≠ unmapped → vols1[v][m] = vols2[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 ∈ vols2) ∧ (v ∈ vols1 → # vols1[v] = # vols2[v] ∧ (∀ m. m < # vols1[v] ∧ vols1[v][m] ≠ unmapped → vols1[v][m] = vols2[v][m])), ∀ m. m < # vols2[v] ∧ vols1[v][m] ≠ unmapped → vols1[v][m] = vols2[v][m], ∀ m. m < # vols2[v] ∧ vols0[v][m] ≠ unmapped → vols0[v][m] = vols1[v][m] ⊦ vols0[v][m] = unmapped, vols0[v][m] = vols2[v][m]

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 11 terms: m

Additional information

Used terms: m