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Proof step 3 from proof and-onesin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Proof step 3 from proof and-onesin (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information