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Proof step 10 from proof get-infto (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m0 < n0, ∀ m, nf. m < n0 → (nf to n0)[m] = nf(m) ⊦ ((λ n. nf(n + 1)) to n0)[m0] = nf(m0 + 1), n0 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2