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Proof step 11 from proof n2s-s2n-identity (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 12:
s→n(n→s(m % 10)) = m % 10, s→n(n→s(m / 10)) = m / 10, 10 ≤ m, ∀ m1. m1 < m → true → s→n(n→s(m1)) = m1 ⊦ m = 0, m / 10 * 10 + s→n(n→s(m % 10)) = m

Rule argument

subst: [m1] → [m % 10]

Additional information

Used simplifier rules: n2s-rec: ⊦ 10 ≤ n → n→s(n) = n→s(n / 10) + n→s(n % 10);%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%mod-le-03: n ≠ 0 ⊦ ¬ n ≤ m % n; from specification nat-div Used terms: m % 10