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Proof step 2 from proof get-infto (previous step)

m < n0 +1, ∀ m, nf. m < n0 → (nf to n0)[m] = nf(m) ⊦ (nf to n0 +1)[m] = nf(m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m ≤ n0, ∀ m, nf. m < n0 → (nf to n0)[m] = nf(m) ⊦ (nf to n0 + 1)[m] = nf(m)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m +1 = m + 1; (s)