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Proof step 1 from proof eq-add2n-1-eq-le (conclusion)

⊦ n = n0 + n1 + 1 ∧ n0 + n1 = m0 + m1 → (n < m0 ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n0 + n1 = m0 + m1, n0 + n1 + 1 < m0 ⊦ m0 = 0

Rule argument

Additional information

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