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Proof step 5 from proof mult-eq-succ (previous step)

m + 1 = m1 * m0, m < m1, m1 = 0 ∨ m0 ≤ 1 ⊦ m0 = 1

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 3

Additional information

Used simplifier rules: zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic