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Proof step 147 from proof bitsor-bi-sgn (previous step)

bitsor(sgn(boolvar1 ' + bits0), bits3) = [], bits2int([]) = bits2int(bitsor(boolvar1 ' + bits0, bits3)), (¬ boolvar1) ∈ bits0, ∀ bits, bits1. # bits1 < #(bits3 + false ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits3 = [], bitsor(boolvar1 ' + bits0, bits3) = [], bits2int([]) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 148:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsor(bits, bits0) ≠ []

Rule argument

name: bitsor-empty rotate: F subst: [bits0, bits] → [bits3, sgn(boolvar1 ' + bits0)]

Additional information

Used simplifier rules: ⊦ sgn(boolvar ' + bits) ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: bits3, sgn(boolvar1 ' + bits0)