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Proof step 125 from proof bitsor-bi-sgn (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 126:
⊦ bits0 ≠ [] → bitsor(true ' + bits, bits0) ≠ false ' + bits1

Rule argument

name: bitsor-first rotate: F subst: [bits1, bits0, bits] → [bits2, bits3, bits0]

Additional information

Used terms: bits2, bits3, bits0