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Proof step 6 from proof two-loup-12 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m + n = m0 * 2 ⊦ m0 < n ↔ m < n

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)