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Proof step 2 from proof mk-rest-01 (previous step)

⊦ m = 0, (mklist(a, m) + x).rest = mklist(a, m - 1) + x

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] → [m, m0] flags:

Additional information

Used simplifier rules: mk-plus: ⊦ mklist(a, m + n) = mklist(a, m) + mklist(a, n);%mklist-one: ⊦ mklist(a, 1) = a ';%mk-last: ⊦ mklist(a, n) + a ' = a ' + mklist(a, 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%c-07: ⊦ (a ' + x).rest = x; from specification list%app-01: ⊦ x + y = x + z ↔ y = z; from specification list%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0