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Proof step 5 from proof bitsor-app (previous step)

1 < #(bits + bits0) ∧ 1 < #(bits1 + bits2), # bits0 = # bits2, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → true → # bits3 = # bits2 ∧ bits ≠ [] ∧ bits3 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits3, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits3, bits2) ⊦ bits = [], bits0 = [], bits1 = [], bits2 = [], bitsor((bits + bits0).butlast, (bits1 + bits2).butlast) + ((bits + bits0).last ∨ (bits1 + bits2).last) = bitsor(bits, bits1) + bitsor(bits0, bits2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# bits0 = # bits2, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → true → # bits3 = # bits2 ∧ bits ≠ [] ∧ bits3 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits3, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits3, bits2) ⊦ bits = [], bits0 = [], bits1 = [], bits2 = [], bitsor(bits + bits0.butlast, bits1 + bits2.butlast) + (bits0.last ∨ bits2.last)' = bitsor(bits, bits1) + bitsor(bits0, bits2), # bits + # bits2 = 0, # bits1 + # bits2 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 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;