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Proof step 99 from proof bxor-flip-one-bit (previous step)

int2bits(i) = bits + boolvar ', bits.first ↔ ¬ (bits.rest + (¬ boolvar)').first, 2 < i, 0 < i ⊦ # bits = 1, i = 0, bits = [], (bits + (¬ boolvar)')[# bits, boolvar] = int2bits(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 100:
⊦ n = # bits → (bits + boolvar ')[n, boolvar0] = bits + boolvar0 '

Rule argument

seq: ⊦ n = # bits → (bits + boolvar ')[n, boolvar0] = bits + boolvar0 ' subst: [boolvar0, n, boolvar, bits] → [boolvar, # bits, ¬ boolvar, bits] rotate: F spec: list-del inst: bitlista name: put-last

Additional information

Used terms: boolvar, # bits, ¬ boolvar, bits