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Proof step 86 from proof bxor-flip-one-bit (previous step)

¬(((bits + (¬ boolvar)').first ↔ ¬ (bits + (¬ boolvar)').rest.first) ∧ 1 < #(bits + (¬ boolvar)')), 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 87:
int2bits(i) = bits + boolvar ', 2 < i, 0 < i ⊦ # bits = 1, i = 0, bits = [], bits.first ↔ ¬ (bits.rest + (¬ boolvar)').first, (sgn(bits + (¬ boolvar)')[# bits] ↔ ¬ boolvar) ∧ sgn(bits + (¬ boolvar)')[# bits, boolvar] = int2bits(i)

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m ≠ 1 → (m + n = 1 ↔ m = 0 ∧ n = 1);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits ≠ [] → (bits + bits0).rest = bits.rest + bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);