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Proof step 7 from proof ordered-app-max (previous step)

a < x.last, (max(x) < a0 ⊃ a0;max(x)) < x.last, ordered≤(a0 ' + x), ∀ a1. a1 < x.last ∧ max(x) < x.last → a1 < max(x) ⊦ x = [], a < (max(x) < a0 ⊃ a0;max(x))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
(max(x) < a0 ⊃ a0;max(x)) < x.last, a < x.last, max(x) < a0, ordered≤(a0 ' + x), ∀ a1. a1 < x.last ∧ max(x) < x.last → a1 < max(x) ⊦ x = [], a < a0
Proof step 10:
(max(x) < a0 ⊃ a0;max(x)) < x.last, a < x.last, ordered≤(a0 ' + x), ∀ a1. a1 < x.last ∧ max(x) < x.last → a1 < max(x) ⊦ x = [], a < max(x), max(x) < a0

Rule argument

pos: right 2

Proof step 7 from proof ordered-app-max (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information