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Proof step 2 from proof put-rec-02 (previous step)
| ⊦ n = 0, (a ' + x)[n, b] = a ' + x[n - 1, b] |
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:
put-succ: ⊦ (b ' + x)[n + 1, a] = b + x[n, a];%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%c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim)
Used terms: n, m