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Proof step 65 from proof min-com (previous step)

min(x + a ' + x3) = min((a ' + x3) + x), min(x3 + a0 ' + x) = min(x + x3), a0 < min((a ' + x3) + x), ∀ x1, x2. #(x1 + x2) < #((a + x3) + (a0 + x)) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ x + x3 = [], (a0 < min(x + x3) ⊃ a0;min(x + x3)) = a0, a < min(x + x3)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 66:
¬ a0 < min(x + x3), min(x + a ' + x3) = min((a ' + x3) + x), min(x3 + a0 ' + x) = min(x + x3), a0 < min((a ' + x3) + x), ∀ x1, x2. #(x1 + x2) < #((a + x3) + (a0 + x)) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ min(x + x3) = a0, x + x3 = [], a < min(x + x3)

Rule argument

pos: right 2

Proof step 65 from proof min-com (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information