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Proof step 65 from proof bitsor-bi-sgn (previous step)

bits3 ≠ [] ∧ boolvar1 ' = true ', ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ boolvar1 ↔ boolvar, bits3 = [], bits2int(true ' + (boolvar ∨ boolvar0)') = bits2int(bitsor(boolvar1 ' + bits0, bits3) + (boolvar ∨ boolvar0)'), (¬ boolvar1) ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 66:
∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits3 = [], bits2int(true ' + boolvar0 ') = bits2int(bitsor(true ' + bits0, bits3) + boolvar0 '), false ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;