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Proof step 6 from proof put-in (previous step)

m ≤ # x0, ∀ a, m, b. a ≠ x0[m] ∧ m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0) ⊦ a = (a0 ' + x0)[m], m = 0, a ∈ x0[m - 1, b] ↔ a = b ∨ a ∈ x0, a = a0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
m = m0 + 1, m ≤ # x0, ∀ a, m, b. a ≠ x0[m] ∧ m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0) ⊦ a = (a0 ' + x0)[m], m = 0, a ∈ x0[m0, b] ↔ a = b ∨ a ∈ x0, a = a0

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m1] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m1