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Proof step 2 from proof ebm-order-write-same (previous step)

v ∈ avols0, avols0 ⊆ avols1 ⊦ avols0[v, avols0[v][m, leb]] ⊆ avols1[v, avols1[v][m, leb]]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
v ∈ avols0, ∀ 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]))) ⊦ (v0 ∈ avols0[v, avols0[v][m, leb]] ↔ v0 ∈ avols1[v, avols1[v][m, leb]]) ∧ ( v0 ∈ avols0[v, avols0[v][m, leb]] → # avols0[v, avols0[v][m, leb]][v0] = # avols1[v, avols1[v][m, leb]][v0] ∧ ( m0 < # avols0[v, avols0[v][m, leb]][v0] ∧ avols0[v, avols0[v][m, leb]][v0][m0] ≠ unmapped → (avols0[v, avols0[v][m, leb]][v0][m0] = erased → avols1[v, avols1[v][m, leb]][v0][m0] = erased) ∧ (avols0[v, avols0[v][m, leb]][v0][m0].mapped? → avols1[v, avols1[v][m, leb]][v0][m0] = avols0[v, avols0[v][m, leb]][v0][m0])))

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]))));