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Proof step 4 from proof fill-sgn-02 (previous step)

fillsgn(# bits0, sgn(bits0)) = bits0 ⊦ fillsgn(1 + # bits0, sgn(boolvar ' + bits0)) = boolvar ' + bits0, 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 ≠ [] ∧ (bits0.first ↔ boolvar)), fillsgn(# bits0, sgn(bits0)) = bits0 ⊦ fillsgn(1 + # bits0, sgn(boolvar ' + bits0)) = boolvar ' + bits0, bits0 = []
Proof step 14:
bits0 ≠ [] ∧ (bits0.first ↔ boolvar), fillsgn(# bits0, sgn(bits0)) = bits0 ⊦ fillsgn(1 + # bits0, sgn(bits0)) = boolvar ' + bits0, bits0 = []

Rule argument

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

Additional information

Used terms: bits0, boolvar