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Proof step 79 from proof bxor-flip-one-bit (previous step)

2 < i, 0 < i, 1 < i, i = 1 ∧ ¬ boolvar0 ∨ i / 2 = ~1 ∧ boolvar0 ⊦ 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 case distinction

And obtained the following new goals

Rule argument

pos: left 4

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