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Proof step 11 from proof max-sort-sublist (previous step)

max(x1 + x0) < a, ∀ x0. x1 + x0 = [] ∨ max(x1 + x0) = max(x1) ∨ max(x1 + x0) = max(x0) ⊦ x1 + x0 = [], a = max(x0), a = max(a ' + x1)

User applied rule all left

And obtained the following new goals

Proof step 12:
max(x1 + x0) < a, x1 + x0 = [] ∨ max(x1 + x0) = max(x1) ∨ max(x1 + x0) = max(x0), ∀ x0. x1 + x0 = [] ∨ max(x1 + x0) = max(x1) ∨ max(x1 + x0) = max(x0) ⊦ x1 + x0 = [], a = max(x0), a = max(a ' + x1)

Rule argument

pos: left 2 terms: x0 abort: F discard: F computed: T

Additional information

Used terms: x0