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Proof step 13 from proof bitsor-app (previous step)

# 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 ≠ []

User applied rule cut formula

And obtained the following new goals

Proof step 14:
# 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) ⊦ bits4 ≠ [], bits = [], bits1 = [], bitsor(bits + bits4, bits1 + bits0) + (boolvar ∨ boolvar0)' = bitsor(bits, bits1) + bitsor(bits4 + boolvar ', bits0 + boolvar0 '), bits4 ≠ [] ∧ bits0 ≠ []

Rule argument

fma: bits4 ≠ []

Additional information

Used simplifier rules: ⊦ # [] = 0;% ⊦ # bits = 0 ↔ bits = []; Used formulas: bits4 ≠ []