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Proof step 9 from proof put-at (previous step)

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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m