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Proof step 22 from proof bitsand-app-ones (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
∀ bits0, bits, bits2. # bits2 < # bits1 → true → bitsand(bits + true ' + bits0, false ' + bits2) = bitsand(true ' + bits0, false ' + bits2) ∨ false ∈ bits ⊦ bits1 = [], bitsand(bits + true ' + bits3 + boolvar ', false ' + bits1) = bitsand(true ' + bits3, false ' + bits1.butlast) + (boolvar ∧ bits1.last)', false ∈ bits

Rule argument

Additional information

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