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Proof step 16 from proof three (previous step)

m0 * m1 + m1 + m0 ≤ 2 ⊦ m1 = 0, m0 = 0, 1 = m1, 1 = m0

User applied rule constructor cut

And closed the goal!

Rule argument

var: m1

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%two-02: ⊦ m + m = 2 * m;%one: ⊦ m0 * 1 = m0;%succ: ⊦ (* m0) * m + m = (m0 + 1) * m;%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);%nat freely generated by 0, +1;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%one: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; from specification nat-basic2%le-num-05: n < n0 ⊦ m + (* n) ≤ (* n0) ↔ m ≤ n0 - n; from specification nat%one-01: ⊦ m ≠ 0 → (m + n ≤ 1 ↔ m = 1 ∧ n = 0); from specification nat