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Proof step 68 from proof bxor-flip-one-bit (previous step)

bits = boolvar0 + bits0, int2bits(i) = bits + boolvar ', bits = boolvar0 ', 2 < i, 0 < i, boolvar ∈ int2bits(i) ⊦ i = 0, bits = [], (sgn(bits + (¬ boolvar)')[# int2bits(i) - 1] ↔ ¬ boolvar) ∧ sgn(bits + (¬ boolvar)')[# int2bits(i) - 1, boolvar] = int2bits(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 69:
int2bits(i) = boolvar0 ' + boolvar ', boolvar0 ∈ int2bits(i), boolvar ∈ int2bits(i), 0 < i, 2 < i ⊦ i = 0, (sgn(boolvar0 ' + (¬ boolvar)')[# int2bits(i) - 1] ↔ ¬ boolvar) ∧ sgn(boolvar0 ' + (¬ boolvar)')[# int2bits(i) - 1, boolvar] = int2bits(i)

Rule argument

Additional information

Used simplifier rules: ⊦ [] + bits = bits;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ boolvar + bits = boolvar ' + bits;