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Proof step 5 from proof twols (previous step)

0 + n0 + 1 < 2, ∀ m. m < n0 ∧ m + n0 < 2 → n0 = 0 ∨ m = 0 ∧ n0 = 1 ⊦ n0 < 0, 0 = 0 ∧ n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
1 + n0 < 2, ∀ m. m < n0 ∧ m + n0 < 2 → m = 0 ∧ n0 = 1 ⊦ n0 = 0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%zero-01: ⊦ ¬ m < 0; from specification nat-basic1