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Proof step 6 from proof bitsxor-first (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ # bits0 = # bits2 ∧ bits ≠ [] ∧ bits0 ≠ [] ∧ bits1 ≠ [] → bitsxor(bits + bits0, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits0, bits2)

Rule argument

name: bitsxor-append-same-size rotate: F subst: [bits2, bits1, bits0, bits] → [bits, boolvar0 ', bits1, boolvar ']

Additional information

Used simplifier rules: bitsxor-bit: ⊦ bitsxor(boolvar ', boolvar0 ') = (¬(boolvar ↔ boolvar0))';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ boolvar ' ≠ [];% ⊦ # [] = 0;% ⊦ # bits = 0 ↔ bits = []; Used terms: bits, boolvar0 ', bits1, boolvar '