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Proof step 4 from proof two-loup-09 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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