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Proof step 2 from proof bitsxor-rec-01-01 (previous step)

⊦ bitsxor(bits, true ') = bitsinvert(bits), bits = []

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ bitsxor(bits + boolvar ', true ') = bitsinvert(bits + boolvar)

Rule argument

name: bitsxor-oner subst: [bits, boolvar] → [bits.butlast, bits.last]

Additional information

Used simplifier rules: bits ≠ [] ⊦ bits.butlast + bits.last ' = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar = bits + boolvar '; Used terms: bits.butlast, bits.last