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Proof step 3 from proof bitsand-app-ones (previous step)

true, ∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ 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

Proof step 3 from proof bitsand-app-ones (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information