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

1 + n0 < n, nnf(n0 + 1) ≤ nnf(n0), ∀ n1. nnf(n1) < nnf(n1 + 1) ⊦ n = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n0

Additional information

Used simplifier rules: ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used terms: n0