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Proof step 86 from proof bitsor-bi-sgn (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ boolvar ∧ bits.first ∧ false ∈ bits → bits2int(bits + boolvar ') = bits2int(bits) * 2 + 1 subst: [boolvar, bits] → [true, bitsor(boolvar1 ' + bits0, bits3)] rotate: F spec: int2bits inst: name: bits2int-rec-03

Additional information

Used terms: true, bitsor(boolvar1 ' + bits0, bits3)