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Proof step 6 from proof put-del1-in (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m0] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0