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Proof step 87 from proof bxor-flip-one-bit (previous step)

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)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 88:
bits.rest + (¬ boolvar)' = boolvar0 + bits0 ∧ bits.rest + (¬ boolvar)' ≠ [], int2bits(i) = bits + boolvar ', 2 < i, 0 < i ⊦ # bits = 1, i = 0, bits = [], bits.first ↔ ¬ boolvar0, (sgn(bits + (¬ boolvar)')[# bits] ↔ ¬ boolvar) ∧ sgn(bits + (¬ boolvar)')[# bits, boolvar] = int2bits(i)

Rule argument

seq: ⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0) subst: [bits, boolvar, bits0] → [bits.rest + (¬ boolvar)', boolvar0, bits0] rotate: T spec: list-data inst: bitlista name: elim-1

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0); (elim) Used terms: bits.rest + (¬ boolvar)', boolvar0, bits0