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Proof step 8 from proof two-loup-12 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m + n = m0 * 2 + 1 ⊦ m0 < n ↔ m < n

Rule argument

Additional information

Used simplifier rules: twols-02: ⊦ m ≠ 0 → (m < 2 ↔ m = 1); from specification nat-basic1