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Proof step 64 from proof and-sgn (previous step)

(¬ boolvar) ∈ bits2, ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits0 + true ' = bits0.first ', sgn(bits2 + boolvar ') = sgn(bitsand(bits0, bits2)) + boolvar ', bits2 = [], bits0 = [], bits2 + boolvar ' = bits2.first ', false ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 65:
(¬ boolvar) ∈ bits ⊦ sgn(bits + boolvar ') = sgn(bits) + boolvar '
Proof step 66:
(¬ boolvar) ∈ bits2, (¬ boolvar) ∈ bits2, ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits0 + true ' = bits0.first ', sgn(bits2) + boolvar ' = sgn(bitsand(bits0, bits2)) + boolvar ', bits2 = [], bits0 = [], bits2 + boolvar ' = bits2.first ', false ∈ bits0

Rule argument

seq: (¬ boolvar) ∈ bits ⊦ sgn(bits + boolvar ') = sgn(bits) + boolvar ' subst: [boolvar, bits] → [boolvar, bits2] rotate: F spec: bitopsaux inst: name: sgn-last-diff

Additional information

Used terms: boolvar, bits2