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Proof step 38 from proof max-ordered (previous step)

(max(x2 + x0) < a ⊃ a;max(x2 + x0)) = a, max(x2) < a, a < max(x2 + x1), max(x2) < max(x2 + x1), ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) = max(x2) ∧ ordered≤(x) ∧ x3 ⊑ x → max(x2 + x3) = max(x2) ⊦ x2 + x1 = [], x2 + x0 = [], max(x2 + x1) = a, x2 = []

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 39:
a = a, max(x2) < max(x2 + x1), a < max(x2 + x1), max(x2) < a, max(x2 + x0) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) = max(x2) ∧ ordered≤(x) ∧ x3 ⊑ x → max(x2 + x3) = max(x2) ⊦ x2 + x0 = [], x2 + x1 = [], max(x2 + x1) = a, x2 = []
Proof step 44:
max(x2 + x0) = a, max(x2) < max(x2 + x1), a < max(x2 + x1), max(x2) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) = max(x2) ∧ ordered≤(x) ∧ x3 ⊑ x → max(x2 + x3) = max(x2) ⊦ x2 + x0 = [], x2 + x1 = [], max(x2 + x1) = a, x2 = [], max(x2 + x0) < a

Rule argument

pos: left 1

Proof step 38 from proof max-ordered (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information