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Proof step 2 from proof mod-succ-06 (previous step)

⊦ (m + 1) % n = m % n + 1, (m + 1) % n = 0, n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
n ≠ 0 ⊦ m % n + 1 ≠ n → (m + 1) % n = m % n + 1

Rule argument

spec: <toplevel> name: mod-succ-01 seq: n ≠ 0 ⊦ m % n + 1 ≠ n → (m + 1) % n = m % n + 1 subst: [n, m] → [n, m]

Additional information

Used simplifier rules: mod-succ: ⊦ m % n + 1 = n → (m + 1) % n = 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + 1 ≠ 0; (s) Used terms: n, m