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Proof step 3 from proof modeq (previous step)

m * n + n0 = m0 * n + n1, n0 < n, n1 < n ⊦ m = m0 ∧ n0 = n1

User applied rule structural induction

And obtained the following new goals

Proof step 4:
m1 +1 * n + n0 = m0 * n + n1, n1 < n, n0 < n, ∀ n, n0, m0, n1. m1 * n + n0 = m0 * n + n1 ∧ n0 < n ∧ n1 < n → m1 = m0 ∧ n0 = n1 ⊦ m1 +1 = m0 ∧ n0 = n1

Rule argument

term: m

Additional information

Used simplifier rules: mult-base: ⊦ m * 0 = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%ls-08: ⊦ m * n + n0 < n ↔ m = 0 ∧ n0 < n;%nat freely generated by 0, +1;;% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic