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Proof step 4 from proof ordered-max (previous step)

max(x0) = x0.last, ordered≤(a ' + x0) ⊦ x0 = [], (x0.last < a ⊃ a;x0.last) = x0.last

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
x0.last < a, max(x0) = x0.last, ordered≤(a ' + x0) ⊦ a = x0.last, x0 = []

Rule argument

pos: right 2

Additional information