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Proof step 6 from proof put-at-eq (previous step)

m ≤ # x0, ∀ m, a. m < # x0 → x0[m, a][m] = a ⊦ m = 0, (a0 ' + x0[m - 1, a])[m] = a

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

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