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

avols ⊆ avols0, avols0 ⊆ avols1 ⊦ avols ⊆ avols1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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]))));