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Proof step 66 from proof bxor-flip-one-bit (previous step)

bits = bits.first + [], int2bits(i) = bits + boolvar ', 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 67:
int2bits(i) = bits + boolvar ', bits = bits.first ', 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)

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;