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Proof step 20 from proof bitsor-app (previous step)

1 < #(bits4 + boolvar ') ∧ 1 < #(bits0 + boolvar0 '), # bits4 = # bits0, ∀ bits1, bits, bits2, bits3. # bits3 < #(bits4 + boolvar ') → true → # bits3 = # bits2 ∧ bits ≠ [] ∧ bits3 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits3, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits3, bits2) ⊦ bits = [], bits1 = [], bitsor(bits + bits4, bits1 + bits0) + (boolvar ∨ boolvar0)' = bitsor(bits, bits1) + (bitsor((bits4 + boolvar ').butlast, (bits0 + boolvar0 ').butlast) + ((bits4 + boolvar ').last ∨ (bits0 + boolvar0 ').last))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# bits4 = # bits0, ∀ bits1, bits, bits2, bits3. # bits3 < #(bits4 + boolvar ') → true → # bits3 = # bits2 ∧ bits ≠ [] ∧ bits3 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits3, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits3, bits2) ⊦ bits = [], bits1 = [], bitsor(bits + bits4, bits1 + bits0) = bitsor(bits, bits1) + bitsor(bits4, bits0), bits4 = [], bits0 = []

Rule argument

Additional information

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