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Proof step 46 from proof and-insgn (previous step)

(boolvar ↔ boolvar0) ∨ boolvar ∈ bits2, ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ boolvar ∈ bitsand(bits0, bits2) ∨ (boolvar ↔ boolvar0), bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', bits0 + true ' = (bits0 + true ').first ', false ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 47:
boolvar ∈ bits2, ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ boolvar ↔ boolvar0, bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', bits0 + true ' = (bits0 + true ').first ', boolvar ∈ bitsand(bits0, bits2), false ∈ bits0

Proof step 46 from proof and-insgn (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information