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Proof step 2 from proof vols-le-antisym (previous step)

vols0 ⊆ vols1, vols1 ⊆ vols0 ⊦ vols0 = vols1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ 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. (v ∈ vols0 ↔ v ∈ vols1) ∧ (v ∈ vols0 → # vols0[v] = # vols1[v] ∧ (∀ m. m < # vols0[v] ∧ vols0[v][m] ≠ unmapped → vols0[v][m] = vols1[v][m])) ⊦ vols0 = vols1

Rule argument

Additional information

Used simplifier rules: vols-le: ⊦ vols0 ⊆ vols1 ↔ (∀ 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])));