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Proof step 2 from proof ebm-order-antisym (previous step)

avols0 ⊆ avols1, avols1 ⊆ avols0 ⊦ avols0 = avols1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ v. (v ∈ avols1 ↔ v ∈ avols0) ∧ ( v ∈ avols1 → # avols1[v] = # avols0[v] ∧ (∀ m. m < # avols1[v] ∧ avols1[v][m] ≠ unmapped → (avols1[v][m] = erased → avols0[v][m] = erased) ∧ (avols1[v][m].mapped? → avols0[v][m] = avols1[v][m]))), ∀ v. (v ∈ avols0 ↔ v ∈ avols1) ∧ ( v ∈ avols0 → # avols0[v] = # avols1[v] ∧ (∀ m. m < # avols0[v] ∧ avols0[v][m] ≠ unmapped → (avols0[v][m] = erased → avols1[v][m] = erased) ∧ (avols0[v][m].mapped? → avols1[v][m] = avols0[v][m]))) ⊦ avols0 = avols1

Rule argument

Additional information

Used simplifier rules: ebm-le: ⊦ avols0 ⊆ avols1 ↔ (∀ v. (v ∈ avols0 ↔ v ∈ avols1) ∧ ( v ∈ avols0 → # avols0[v] = # avols1[v] ∧ (∀ m. m < # avols0[v] ∧ avols0[v][m] ≠ unmapped → (avols0[v][m] = erased → avols1[v][m] = erased) ∧ ( avols0[v][m].mapped? → avols1[v][m] = avols0[v][m]))));