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Proof step 83 from proof bxor-flip-one-bit (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 84:
int2bits(i) = bits + boolvar ', 2 < i, 0 < i ⊦ # bits = 1, i = 0, bits = [], (sgn(bits + (¬ boolvar)')[# bits] ↔ ¬ boolvar) ∧ sgn(bits + (¬ boolvar)')[# bits, boolvar] = int2bits(i)

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);