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Proof step 9 from proof get-infto (previous step)

m ≤ n0, ∀ m, nf. m < n0 → (nf to n0)[m] = nf(m) ⊦ ((λ n. nf(n + 1)) to n0)[m - 1] = nf(m), m = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
m = m0 + 1, m ≤ n0, ∀ m, nf. m < n0 → (nf to n0)[m] = nf(m) ⊦ ((λ n. nf(n + 1)) to n0)[m0] = nf(m), m = 0

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m2] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m2