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Proof step 60 from proof bxor-flip-one-bit (previous step)

int2bits(i) = boolvar ', 2 < i, 0 < i ⊦ i = 0, (sgn(bitsxor([], false ') + (¬ boolvar)')[0] ↔ ¬ boolvar) ∧ sgn(bitsxor([], false ') + (¬ boolvar)')[0, boolvar] = boolvar '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 61:
⊦ int2bits(i) = boolvar ' ↔ i = 0 ∧ ¬ boolvar ∨ i = ~1 ∧ boolvar

Rule argument

seq: ⊦ int2bits(i) = boolvar ' ↔ i = 0 ∧ ¬ boolvar ∨ i = ~1 ∧ boolvar subst: [boolvar, i] → [boolvar, i] rotate: F spec: int2bits inst: name: int2bits-one-bit

Additional information

Used terms: boolvar, i