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Proof step 143 from proof bxor-flip-one-bit (previous step)

int2bits(i) = (bits1 + boolvar ') + bits, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))), 0 ≤ i, 0 ≤ j, 3 + i→n(j) ≤ # int2bits(i), 0 < i ⊦ bits = [], bits1 = [], bits1.first ↔ ¬ (bits1.rest + (¬ boolvar)').first, i = 2 ^ j * 2, (sgn((bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1] ↔ ¬ boolvar) ∧ sgn((bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1, boolvar] = int2bits(i)

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: ⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0) subst: [bits, boolvar, bits0] → [bits1, boolvar0, bits0] rotate: T spec: list-data inst: bitlista name: elim-1

Additional information

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