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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 ⊦ (a0 ' + x0[m - 1, a])[m] = a, m = 0

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 ⊦ (a0 ' + x0[m0, a])[m] = a, m = 0

Rule argument

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

Additional information

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