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Proof step 56 from proof bxor-flip-one-bit (previous step)

int2bits(i) = bits + boolvar ' ∧ int2bits(i) ≠ [], 2 < i, 0 < 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 57:
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)

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' = bits0 → boolvar ∈ bits0; (forward)% ⊦ int2bits(j) ≠ [];