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Proof step 126 from proof bxor-flip-one-bit (previous step)

int2bits(i) = false ' + bits, # int2bits(i) = 2 + i→n(j), # bits = i→n(j), bitsxor(int2bits(i).rest, true ' + mklist(false, i→n(j))) = true ' + bitsxor(bits, mklist(false, i→n(j))), 0 ≤ j, false ∈ int2bits(i), 2 ^ j * 2 < i, 0 < i ⊦ bits = [], i = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 127:
int2bits(i) = boolvar + bits0 ∧ int2bits(i) ≠ [], int2bits(i) = false ' + bits, # int2bits(i) = 2 + i→n(j), # bits = i→n(j), bitsxor(bits0, true ' + mklist(false, i→n(j))) = true ' + bitsxor(bits, mklist(false, i→n(j))), 0 ≤ j, false ∈ int2bits(i), 2 ^ j * 2 < i, 0 < i ⊦ bits = [], i = 0

Rule argument

seq: ⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0) subst: [bits, boolvar, bits0] → [int2bits(i), boolvar, bits0] rotate: T spec: list-data inst: bitlista name: elim-1

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];%⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0); (elim) Used terms: int2bits(i), boolvar, bits0