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Proof step 3 from proof bitsand-onein (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ bits0, bits, bits2. # bits2 < # bits1 → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ true ∈ bitsand(bits + false ' + bits0, bits1) ↔ true ∈ bitsand(false ' + bits0, bits1), bits1 = [], true ∈ bits

Proof step 3 from proof bitsand-onein (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information