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Proof step 4 from proof sgn-append-03 (previous step)

bits0 = [] ∨ (∀ bits1, boolvar. sgn(bits0 + bits1) = sgn((¬ boolvar)' + bits1) ∨ boolvar ∈ bits0) ⊦ sgn((boolvar ' + bits0) + bits1) = sgn((¬ boolvar0)' + bits1), boolvar0 ↔ boolvar, boolvar0 ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
bits ≠ [], bits.first ↔ boolvar ⊦ sgn(boolvar ' + bits) = sgn(bits)
Proof step 6:
¬(bits0 + bits1 ≠ [] ∧ ((bits0 + bits1).first ↔ boolvar)), bits0 = [] ∨ (∀ bits1, boolvar. sgn(bits0 + bits1) = sgn((¬ boolvar)' + bits1) ∨ boolvar ∈ bits0) ⊦ sgn((boolvar ' + bits0) + bits1) = sgn((¬ boolvar0)' + bits1), boolvar0 ↔ boolvar, boolvar0 ∈ bits0
Proof step 15:
bits0 + bits1 ≠ [] ∧ ((bits0 + bits1).first ↔ boolvar), bits0 = [] ∨ (∀ bits1, boolvar. sgn(bits0 + bits1) = sgn((¬ boolvar)' + bits1) ∨ boolvar ∈ bits0) ⊦ sgn(bits0 + bits1) = sgn((¬ boolvar0)' + bits1), boolvar0 ↔ boolvar, boolvar0 ∈ bits0

Rule argument

name: sgn-rec-01 rotate: F subst: [bits, boolvar] → [bits0 + bits1, boolvar]

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: bits0 + bits1, boolvar