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Proof step 23 from proof sgn-ex-01 (previous step)

true ∈ bits0 ⊦ false ∈ bits0, ∃ boolvar2, n0, boolvar3, bits1. false ' + bits0 = mklist(boolvar2, n0) + boolvar3 ' + bits1 ∧ sgn(mklist(boolvar2, n0) + boolvar3 ' + bits1) = boolvar2 ' + boolvar3 ' + bits1 ∧ ¬(boolvar2 ↔ boolvar3) ∧ n0 ≠ 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 24:
⊦ ¬ boolvar ∈ bits ↔ mklist(¬ boolvar, # bits) = bits
Proof step 25:
true ∈ bits0 ⊦ mklist(¬ false, # bits0) ≠ bits0, ∃ boolvar2, n0, boolvar3, bits1. false ' + bits0 = mklist(boolvar2, n0) + boolvar3 ' + bits1 ∧ sgn(mklist(boolvar2, n0) + boolvar3 ' + bits1) = boolvar2 ' + boolvar3 ' + bits1 ∧ ¬(boolvar2 ↔ boolvar3) ∧ n0 ≠ 0

Rule argument

seq: ⊦ ¬ boolvar ∈ bits ↔ mklist(¬ boolvar, # bits) = bits subst: [bits, boolvar] → [bits0, false] rotate: F spec: bitlista inst: name: mklist-intro

Additional information

Used terms: bits0, false