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Proof step 62 from proof bitsor-bi-sgn (previous step)

bits3 ≠ [] ∧ false ' = false ', ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(bits3 + true ') = bits2int(bitsor(false ' + bits0, bits3) + true '), bits3 = [], true ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 63:
∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(bits3 + true ') = bits2int(bitsor(false ' + bits0, bits3) + true '), bits3 = [], true ∈ bits0

Proof step 62 from proof bitsor-bi-sgn (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information