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Proof step 6 from proof ins-01 (previous step)

a0 < a, ∀ a, y. sort(ins≤(a, x0) + y) = ins≤(a, sort(x0 + y)) ⊦ ins≤(a0, sort(ins≤(a, x0) + y)) = ins≤(a, ins≤(a0, sort(x0 + y)))

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a, y abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ ins≤(a, ins≤(b, x)) = ins≤(b, ins≤(a, x)) ↔ true; (s) Used terms: a, y