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Proof step 3 from proof sgn-sgn-01 (previous step)

∀ bits1. sgn(sgn(bits0) + bits1) = sgn(bits0 + bits1) ⊦ sgn(sgn(boolvar ' + bits0) + bits1) = sgn((boolvar ' + bits0) + bits1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
bits ≠ [], bits.first ↔ boolvar ⊦ sgn(boolvar ' + bits) = sgn(bits)
Proof step 5:
¬(bits0 ≠ [] ∧ (bits0.first ↔ boolvar)), ∀ bits1. sgn(sgn(bits0) + bits1) = sgn(bits0 + bits1) ⊦ sgn(sgn(boolvar ' + bits0) + bits1) = sgn((boolvar ' + bits0) + bits1)
Proof step 9:
bits0 ≠ [] ∧ (bits0.first ↔ boolvar), ∀ bits1. sgn(sgn(bits0) + bits1) = sgn(bits0 + bits1) ⊦ sgn(sgn(bits0) + bits1) = sgn((boolvar ' + bits0) + bits1)

Rule argument

name: sgn-rec-01 rotate: F subst: [bits, boolvar] → [bits0, boolvar]

Additional information

Used terms: bits0, boolvar