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

∀ boolvar. sgn(boolvar ' + bits0).first ↔ boolvar ⊦ sgn(boolvar0 ' + boolvar ' + bits0).first ↔ boolvar0

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:
¬(boolvar ' + bits0 ≠ [] ∧ ((boolvar ' + bits0).first ↔ boolvar0)), ∀ boolvar. sgn(boolvar ' + bits0).first ↔ boolvar ⊦ sgn(boolvar0 ' + boolvar ' + bits0).first ↔ boolvar0
Proof step 7:
boolvar ' + bits0 ≠ [] ∧ ((boolvar ' + bits0).first ↔ boolvar0), ∀ boolvar. sgn(boolvar ' + bits0).first ↔ boolvar ⊦ sgn(boolvar ' + bits0).first ↔ boolvar0

Rule argument

name: sgn-rec-01 rotate: F subst: [bits, boolvar] → [boolvar ' + bits0, boolvar0]

Additional information

Used terms: boolvar ' + bits0, boolvar0