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Proof step 23 from proof min-com (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 24:
a0 < min(x + x3), min(x3 + a0 ' + x) = a0, min(x + a ' + x3) = min((a ' + x3) + x), a0 < min((a ' + x3) + x), a < (a0 < min(x + x3) ⊃ a0;min(x + x3)), ∀ x1, x2. #(x1 + x2) < #((a + x3) + (a0 + x)) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ a = a0, x + x3 = []
Proof step 27:
¬ a0 < min(x + x3), min(x3 + a0 ' + x) = min(x + x3), min(x + a ' + x3) = min((a ' + x3) + x), a0 < min((a ' + x3) + x), a < (a0 < min(x + x3) ⊃ a0;min(x + x3)), ∀ x1, x2. #(x1 + x2) < #((a + x3) + (a0 + x)) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ a = a0, x + x3 = []

Rule argument

pos: left 2

Proof step 23 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