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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information