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Proof step 68 from proof bitsor-bi-sgn (previous step)

(¬ bits2.first) ∈ bits2.rest ∧ bits2 ≠ [], ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(bitsor(sgn(bits2) + boolvar ', bits3 + boolvar0 ')) = bits2int(bitsor(bits2, bits3) + (boolvar ∨ boolvar0)'), bits2 = [], bits3 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 69:
(¬ bits2.first) ∈ bits2.rest, ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2 = [], bits3 = [], bits2int(bitsor(sgn(bits2) + boolvar ', bits3 + boolvar0 ')) = bits2int(bitsor(bits2, bits3) + (boolvar ∨ boolvar0)')

Proof step 68 from proof bitsor-bi-sgn (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information