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Proof step 153 from proof bxor-flip-one-bit (previous step)

int2bits(i) = (false ' + bits0) + boolvar ' + bits, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (false ' + bits0) + (¬ boolvar)' + bitsxor(bits, mklist(false, i→n(j))), 0 < i, 0 ≤ j, 3 + i→n(j) ≤ # int2bits(i), false ∈ int2bits(i), false ∈ bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) ⊦ bits0 = [], i = 0, # int2bits(i) = 2 + i→n(j), i = 2 ^ j * 2, bits = [], bits0.first, (sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[1 + # bits0] ↔ ¬ boolvar) ∧ sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[1 + # bits0, 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] → [bits0, boolvar0, bits1] 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: bits0, boolvar0, bits1