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Proof step 6 from proof n2s-s2n-identity (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ str ≠ "" ∧ str0 ≠ "" → s→n(str + str0) = s→n(str) * 10 ^ # str0 + s→n(str0)
Proof step 8:
n→s(m / 10) ≠ "" ∧ n→s(m % 10) ≠ "", 10 ≤ m, ∀ m1. m1 < m → true → s→n(n→s(m1)) = m1 ⊦ s→n(n→s(m / 10)) * 10 ^ # n→s(m % 10) + s→n(n→s(m % 10)) = m, m = 0

Rule argument

spec: inst: name: s2n-app seq: ⊦ str ≠ "" ∧ str0 ≠ "" → s→n(str + str0) = s→n(str) * 10 ^ # str0 + s→n(str0) subst: [str0, str] → [n→s(m % 10), n→s(m / 10)] flags: rewrite, current

Additional information

Used simplifier rules: n2s-empty: ⊦ n→s(n) ≠ ""; Used terms: n→s(m % 10), n→s(m / 10)