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Proof step 4 from proof fst-nle (previous step)

m.1 ≤ m, m0 ≠ 0, m0 ≠ 1 ⊦ n + m * m0 ≤ m.1 ↔ m = 0 ∧ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m.1 ≤ m, n + m * m0 ≤ m.1, n + m * m0 ≤ m ⊦ m0 = 1, m0 = 0, m = 0 ∧ n = 0

Rule argument

Additional information

Used simplifier rules: fst-zero: ⊦ 0.1 = 0;%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%zero: ⊦ 0 * n = 0; from specification nat-mult