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Proof step 5 from proof n2s-s2n-identity (previous step)

¬ m < 10, ∀ m1. m1 < m → true → s→n(n→s(m1)) = m1 ⊦ s→n(n→s(m)) = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
10 ≤ m, ∀ m1. m1 < m → true → s→n(n→s(m1)) = m1 ⊦ s→n(n→s(m / 10) + n→s(m % 10)) = m, m = 0

Rule argument

Additional information

Used simplifier rules: n2s-rec: ⊦ 10 ≤ n → n→s(n) = n→s(n / 10) + n→s(n % 10);%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat