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Proof step 8 from proof bitsand-app (previous step)

bits1 = bits4 + boolvar ', # bits = # bits1, ∀ bits2, bits0, bits, bits3. # bits3 < # bits1 → # bits = # bits3 → bits0 = [] ∨ bits2 = [] ∨ bits = [] ∨ bitsand(bits0 + bits, bits2 + bits3) = bitsand(bits0, bits2) + bitsand(bits, bits3) ⊦ # bits0 + # bits1 = 0, # bits2 + # bits1 = 0, bits1 = [], bits = [], bits2 = [], bits0 = [], bitsand(bits0 + bits.butlast, bits2 + bits4) + (bits.last ∧ boolvar)' = bitsand(bits0, bits2) + bitsand(bits, bits1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# bits = # bits4 + 1, ∀ bits2, bits0, bits, bits3. # bits3 < #(bits4 + boolvar ') → # bits = # bits3 → bits0 = [] ∨ bits2 = [] ∨ bits = [] ∨ bitsand(bits0 + bits, bits2 + bits3) = bitsand(bits0, bits2) + bitsand(bits, bits3) ⊦ bits = [], bits2 = [], bits0 = [], bitsand(bits0 + bits.butlast, bits2 + bits4) + (bits.last ∧ boolvar)' = bitsand(bits0, bits2) + bitsand(bits, bits4 + boolvar ')

Rule argument

Additional information

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