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Proof step 10 from proof min-com (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ x1, x2. #(x1 + x2) < #((a + x3) + (a0 + x)) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ (a < min(x3 + a0 ' + x) ⊃ a;min(x3 + a0 ' + x)) = min((a0 ' + x) + a ' + x3)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%e-03: ⊦ x + a ' + y ≠ []; from specification list