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Proof step 95 from proof min-com (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 96:
a < min(x3 + x), min(x3 + a0 ' + x) = min(x + x3), min(x + a ' + x3) = (a < min(x3 + x) ⊃ a;min(x3 + x)), ∀ x1, x2. #(x1 + x2) < #((a + x3) + (a0 + x)) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ min(x + x3) = a, x + x3 = [], x3 + x = [], a < min(x + x3), a0 < min(x + x3), a0 < (a < min(x3 + x) ⊃ a;min(x3 + x))
Proof step 98:
¬ a < min(x3 + x), min(x3 + a0 ' + x) = min(x + x3), min(x + a ' + x3) = (a < min(x3 + x) ⊃ a;min(x3 + x)), ∀ x1, x2. #(x1 + x2) < #((a + x3) + (a0 + x)) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ min(x + x3) = min(x3 + x), x + x3 = [], x3 + x = [], a < min(x + x3), a0 < min(x + x3), a0 < (a < min(x3 + x) ⊃ a;min(x3 + x))

Rule argument

pos: right 2

Proof step 95 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