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Proof step 10 from proof put-at (previous step)

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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

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