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Proof step 4 from proof two-loup-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1