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Proof step 80 from proof bxor-flip-one-bit (previous step)

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)'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 81:
i / 2 = ~1, 2 < i, 0 < i, 1 < i ⊦ i = 0, (i % 2 = 0 → true '[1]) ∧ sgn(true ' + (i % 2 = 0)')[1, i % 2 ≠ 0] = true ' + (i % 2 ≠ 0)'

Rule argument

Additional information

Used simplifier rules: ⊦ sgn(true ' + true ') = true ';% ⊦ boolvar '[0] ↔ boolvar;%n ≠ 0 ⊦ (boolvar ' + bits)[(* n)] ↔ bits[n -1];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ sgn(true ' + false ') = true ' + false ';