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Proof step 76 from proof bxor-flip-one-bit (previous step)

int2bits(i / 2) = boolvar0 ', 2 < i, 0 < i, 1 < i ⊦ i = 0, (sgn(boolvar0 ' + (i % 2 = 0)')[1] ↔ i % 2 = 0) ∧ sgn(boolvar0 ' + (i % 2 = 0)')[1, i % 2 ≠ 0] = boolvar0 ' + (i % 2 ≠ 0)'

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 77:
⊦ int2bits(i) = boolvar ' ↔ i = 0 ∧ ¬ boolvar ∨ i = ~1 ∧ boolvar
Proof step 78:
i / 2 = 0 ∧ ¬ boolvar0 ∨ i / 2 = ~1 ∧ boolvar0, 2 < i, 0 < i, 1 < i ⊦ i = 0, (sgn(boolvar0 ' + (i % 2 = 0)')[1] ↔ i % 2 = 0) ∧ sgn(boolvar0 ' + (i % 2 = 0)')[1, i % 2 ≠ 0] = boolvar0 ' + (i % 2 ≠ 0)'

Rule argument

seq: ⊦ int2bits(i) = boolvar ' ↔ i = 0 ∧ ¬ boolvar ∨ i = ~1 ∧ boolvar subst: [boolvar, i] → [boolvar0, i / 2] rotate: F spec: int2bits inst: name: int2bits-one-bit

Additional information

Used terms: boolvar0, i / 2