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Proof step 14 from proof bitsxor-append-same-size (previous step)

1 < # bits0 ∧ 1 < # bits2, # bits0 = # bits2, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → # bits3 = # bits2 → bits = [] ∨ bits3 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits3, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits3, bits2) ⊦ # bits + # bits2 = 0, # bits1 + # bits2 = 0, bits2 = [], bits1 = [], bits0 = [], bits = [], bitsxor(bits + bits0.butlast, bits1 + bits2.butlast) + (¬(bits0.last ↔ bits2.last))' = bitsxor(bits, bits1) + (bitsxor(bits0.butlast, bits2.butlast) + (¬(bits0.last ↔ bits2.last)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# bits0 = # bits2, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → # bits3 = # bits2 → bits = [] ∨ bits3 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits3, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits3, bits2) ⊦ # bits2 = 1, bitsxor(bits + bits0.butlast, bits1 + bits2.butlast) = bitsxor(bits, bits1) + bitsxor(bits0.butlast, bits2.butlast), bits = [], bits0 = [], bits1 = [], bits2 = [], # bits1 + # bits2 = 0, # bits + # bits2 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits1 + bits0 ↔ bits = bits1;% ⊦ bits + boolvar = bits + boolvar ';