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Proof step 4 from proof sgn-last-04 (previous step)

∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ sgn((boolvar ' + bits0) + boolvar0 ') = sgn(boolvar ' + bits0) + boolvar0 ', ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar)

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 + boolvar0 ' ≠ [] ∧ ((bits0 + boolvar0 ').first ↔ boolvar)), ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ sgn((boolvar ' + bits0) + boolvar0 ') = sgn(boolvar ' + bits0) + boolvar0 ', ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar)
Proof step 16:
bits0 + boolvar0 ' ≠ [] ∧ ((bits0 + boolvar0 ').first ↔ boolvar), ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ sgn(bits0 + boolvar0 ') = sgn(boolvar ' + bits0) + boolvar0 ', ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar)

Rule argument

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

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: bits0 + boolvar0 ', boolvar