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Proof step 8 from proof monls-from (previous step)

nnf((n0 + 1) + n) ≤ nnf(n0 + n), ∀ n1. nnf(n1) < nnf(n1 + 1) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: n0 + n

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used terms: n0 + n