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Proof step 6 from proof bitsxor-append-same-size (previous step)

1 < #(bits + bits0) ∧ 1 < #(bits1 + 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) ⊦ bitsxor((bits + bits0).butlast, (bits1 + bits2).butlast) + (¬((bits + bits0).last ↔ (bits1 + bits2).last)) = bitsxor(bits, bits1) + bitsxor(bits0, bits2), bits1 = [], bits0 = [], bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# 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, bits2)

Rule argument

Additional information

Used simplifier rules: ⊦ # [] = 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ bits ≠ [] → ((bits0 + bits).last ↔ bits.last);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → (bits0 + bits).butlast = bits0 + bits.butlast;