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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ bits0, bits, bits2. # bits2 < # [] → true → bitsand(bits + true ' + bits0, false ' + bits2) = bitsand(true ' + bits0, false ' + bits2) ∨ false ∈ bits ⊦ bits0 = [], bitsand(bits + true ' + bits0, false ') = bitsand(true ' + bits0, false '), false ∈ bits

Rule argument

Additional information

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