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Proof step 13 from proof bitsxor-append-same-size (previous step)

bits2 = bits0 + boolvar0 ', # bits2 = 1, ∀ bits1, bits, bits2, bits3. # bits3 < 1 → # bits3 = # bits2 → bits = [] ∨ bits3 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits3, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits3, bits2) ⊦ bits = [], bits1 = [], bits2 = [], bitsxor(bits, bits1 + bits0) + (¬(boolvar ↔ boolvar0))' = bitsxor(bits, bits1) + bitsxor(boolvar ', bits2)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: bitsxor-bit: ⊦ bitsxor(boolvar ', boolvar0 ') = (¬(boolvar ↔ boolvar0))';% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ [] + bits = bits;% ⊦ bits + [] = bits;% ⊦ bits + boolvar ' ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;