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Proof step 122 from proof bitsor-bi-sgn (previous step)

bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3)) = bits2uint(bits2), bitsor(boolvar1 ' + bits0, bits3) = false ' + bits2, (¬ boolvar1) ∈ bits0, boolvar ∨ boolvar0, ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ ¬ bitsor(sgn(boolvar1 ' + bits0), bits3).first, bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3) + true ') = 1 + 2 * bits2uint(bits2), bits3 = [], ¬ boolvar ∧ ¬ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 123:
bitsor(boolvar1 ' + bits0, bits3) = false ' + bits2, bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3)) = bits2uint(bits2), (¬ boolvar1) ∈ bits0, bitsor(sgn(boolvar1 ' + bits0), bits3).first, boolvar ∨ boolvar0, ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits3 = [], bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3) + true ') = 1 + 2 * bits2uint(bits2), ¬ boolvar ∧ ¬ boolvar0

Proof step 122 from proof bitsor-bi-sgn (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information