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Proof step 6 from proof bitsxor-rec-01 (previous step)

⊦ bitsxor(bits1 + boolvar ', bits + boolvar0 ') = bitsxor(bits1, bits) + (¬(boolvar ↔ boolvar0))', bits1 = [], bits = []

User applied rule insert lemma

And obtained the following new goals

Proof step 7:
⊦ bitsxor(bits + boolvar ' + boolvar0 ', bits0 + boolvar1 ' + boolvar2 ') = bitsxor(bits + boolvar, bits0 + boolvar1) + (¬(boolvar0 ↔ boolvar2))

Rule argument

name: bitsxor-rec subst: [bits, boolvar, boolvar0, bits0, boolvar1, boolvar2] → [bits1.butlast, bits1.last, boolvar, bits.butlast, bits.last, boolvar0]

Additional information

Used simplifier rules: bits ≠ [] ⊦ bits.butlast + bits.last ' = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar = bits + boolvar '; Used terms: bits1.butlast, bits1.last, boolvar, bits.butlast, bits.last, boolvar0