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Proof step 6 from proof twols (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: 1

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1 Used terms: 1