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Proof step 9 from proof n2s-s2n-identity (previous step)

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

User applied rule apply induction

And obtained the following new goals

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

Rule argument

subst: [m1] → [m / 10]

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%le-div-to-mult: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m; from specification nat-div%leone: ⊦ n ≠ 0 → (m * n ≤ n ↔ m ≤ 1); from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: m / 10