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Proof step 39 from proof max-ordered (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 40:
max(x2 + x0) < a, max(x2) < a, ordered≤(x0), x1 ⊑ x0, a < max(x2 + x1), max(x2) < max(x2 + x1), max(x2 + x0) < max(x2 + x1), max(x2 + x0) = max(x2) ∧ ordered≤(x0) ∧ x1 ⊑ x0 → max(x2 + x1) = max(x2) ⊦ x2 + x0 = [], max(x2 + x1) = a, x2 + x1 = [], x2 = []

Rule argument

pos: left 8 terms: x0, x1 abort: F discard: T computed: T

Additional information

Used terms: x0, x1