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Proof step 11 from proof put-at (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
m = m0 + 1, m ≤ # x0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ m = 0, m = n, (a0 ' + x0[m0, a])[n] = (a0 ' + x0)[n]

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