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Proof step 4 from proof sort-max-last (previous step)

∀ a, x0. sort(x1 + a ' + x0) = sort(x1 + x0) + a ' ∨ a < max(x1 + x0) ⊦ ins≤(a0, sort(x1 + a ' + x0)) = ins≤(a0, sort(x1 + x0)) + a ', a < max((a0 ' + x1) + x0)

User applied rule all left

And obtained the following new goals

Proof step 5:
sort(x1 + a ' + x0) = sort(x1 + x0) + a ' ∨ a < max(x1 + x0) ⊦ ins≤(a0, sort(x1 + a ' + x0)) = ins≤(a0, sort(x1 + x0)) + a ', a < max((a0 ' + x1) + x0)

Rule argument

pos: left 1 terms: a, x0 abort: F discard: T computed: F

Additional information

Used terms: a, x0