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Proof step 5 from proof sort-max-last (previous step)

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)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
a < max(x1 + 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)
Proof step 9:
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), a < max(x1 + x0)

Rule argument

fma: a < max(x1 + x0)

Additional information

Used formulas: a < max(x1 + x0)