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Proof step 3 from proof vols-le-trans (previous step)

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

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 4:
v ∈ vols1 ↔ v ∈ vols2, v ∈ vols1 → # vols1[v] = # vols2[v] ∧ (∀ m. m < # vols1[v] ∧ 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])) ⊦ (v ∈ vols0 ↔ v ∈ vols2) ∧ (v ∈ vols0 → # vols0[v] = # vols2[v] ∧ (m < # vols0[v] ∧ vols0[v][m] ≠ unmapped → vols0[v][m] = vols2[v][m]))

Rule argument

pos: left 1 terms: v

Additional information

Used terms: v