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Proof step 10 from proof ebm-order-write-same (previous step)

v ∈ avols0 ↔ v ∈ avols1, v ∈ avols0, 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])) ⊦ m0 = m, # avols0[v] = # avols1[v] ∧ ( m0 < # avols0[v] ∧ avols0[v][m0] ≠ unmapped → (avols0[v][m0] = erased → avols1[v][m0] = erased) ∧ (avols0[v][m0].mapped? → avols1[v][m0] = avols0[v][m0]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# avols0[v] = # avols1[v], m0 < # avols1[v], v ∈ avols1, v ∈ avols0, ∀ m. m < # avols1[v] ∧ avols0[v][m] ≠ unmapped → (avols0[v][m] = erased → avols1[v][m] = erased) ∧ (avols0[v][m].mapped? → avols1[v][m] = avols0[v][m]) ⊦ avols0[v][m0] = unmapped, m0 = m, (avols0[v][m0] = erased → avols1[v][m0] = erased) ∧ (avols0[v][m0].mapped? → avols1[v][m0] = avols0[v][m0])

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic