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Proof step 4 from proof two-loup-05 (previous step)

m + n = m0 * 2 + n0 ∧ n0 < 2 ∧ 2 ≠ 0, m0 + 1 < n, m < n, m0 + 1 < m, 2 + m0 < n ⊦ n = 0, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m + n = m0 * 2 + n0, m0 + 1 < n, m0 + 1 < m, m < n, n0 < 2, 2 + m0 < n ⊦ m = 0, n = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1