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Proof step 6 from proof del1 (previous step)

∀ a. sort(x0 -1l a) = sort(x0) -1l a ⊦ a = a0, ins≤(a0, sort(x0 -1l a)) = ins≤(a0, sort(x0)) -1l a

User applied rule all left

And obtained the following new goals

Proof step 7:
sort(x0 -1l a) = sort(x0) -1l a ⊦ a = a0, ins≤(a0, sort(x0 -1l a)) = ins≤(a0, sort(x0)) -1l a

Rule argument

pos: left 1 terms: a abort: F discard: T computed: F

Additional information

Used terms: a