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Proof step 59 from proof bxor-flip-one-bit (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
int2bits(i) = boolvar ', 2 < i, 0 < i ⊦ i = 0, (sgn(bitsxor([], false ') + (¬ boolvar)')[0] ↔ ¬ boolvar) ∧ sgn(bitsxor([], false ') + (¬ boolvar)')[0, boolvar] = boolvar '

Rule argument

Additional information

Used simplifier rules: ⊦ # boolvar ' = 0 +1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;