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Proof step 14 from proof max-sort-sublist (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 15:
max(x1 + x0) = max(x0), max(x1 + x0) < a, ∀ x0. x1 + x0 = [] ∨ max(x1 + x0) = max(x0) ∨ max(x1 + x0) = max(x1) ⊦ x1 + x0 = [], a = max(a ' + x1), a = max(x0)
Proof step 20:
max(x1 + x0) = max(x1), max(x1 + x0) < a, ∀ x0. x1 + x0 = [] ∨ max(x1 + x0) = max(x0) ∨ max(x1 + x0) = max(x1) ⊦ x1 + x0 = [], a = max(a ' + x1), a = max(x0)

Rule argument

pos: left 3

Proof step 14 from proof max-sort-sublist (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information