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Proof step 30 from proof orderedls-all (previous step)

m1 < # y, m ≤ # y, ordered<(a0 ' + y), a < a0, m1 < m, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ y = [], (a0 ' + y)[m1] < (a0 ' + y)[m]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: m1, m

Additional information

Used simplifier rules: le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used terms: m1, m