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Proof step 10 from proof l (previous step)

m0 + n < m0 + n0, ∀ m, n1, n2. ¬ m0 < m → (m + n1 < m + n2 ↔ n1 < n2) ⊦ m0 + n0 = 0, n < n0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: m0, n, n0

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%l-04: ⊦ m + m0 < n + n0 → (¬ m + m0 < n + n0 ↔ false);%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: m0, n, n0