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Proof step 12 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 ', bits0 + boolvar0 ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# 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 ', bits0 + boolvar0 '), bits4 ≠ [] ∧ bits0 ≠ []

Rule argument

Additional information

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