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Proof step 7 from proof bitsor-app (previous step)

bits0 = bits4 + boolvar ', # 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 + bits4, bits1 + bits2.butlast) + (boolvar ∨ bits2.last)' = bitsor(bits, bits1) + bitsor(bits0, bits2), # bits + # bits2 = 0, # bits1 + # bits2 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# bits4 + 1 = # bits2, ∀ 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 = [], bits2 = [], bitsor(bits + bits4, bits1 + bits2.butlast) + (boolvar ∨ bits2.last)' = bitsor(bits, bits1) + bitsor(bits4 + boolvar ', bits2), # bits + # bits2 = 0, # bits1 + # bits2 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;