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Proof step 6 from proof two-loup-07 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m + n = m0 * 2 + n0, n0 < 2, n + 1 < m0, m < n, 2 * m0 ≤ m + n ⊦ m0 = 0, n = 0

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat