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Proof step 45 from proof bitsand-app-zeros-01 (previous step)

¬(true ∈ bitsand(bits + false ' + bits3, bits4) ∧ false ∈ bitsand(bits + false ' + bits3, bits4)), false ∈ bitsand(false ' + bits3, bits4), true ∈ bitsand(false ' + bits3, bits4), ∀ bits0, bits, bits2. # bits2 < #(bits4 + boolvar0 ') → true → bits2 = [] ∨ sgn(bitsand(bits + false ' + bits0, bits2)) = sgn(bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ sgn(bitsand(bits + false ' + bits3, bits4) + (boolvar ∧ boolvar0)') = sgn(bitsand(false ' + bits3, bits4)) + (boolvar ∧ boolvar0)', bits4 = [], true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
false ∈ bitsand(false ' + bits3, bits4), true ∈ bitsand(false ' + bits3, bits4), ∀ bits0, bits, bits2. # bits2 < #(bits4 + boolvar0 ') → true → bits2 = [] ∨ sgn(bitsand(bits + false ' + bits0, bits2)) = sgn(bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ sgn(bitsand(bits + false ' + bits3, bits4) + (boolvar ∧ boolvar0)') = sgn(bitsand(false ' + bits3, bits4)) + (boolvar ∧ boolvar0)', bits4 = [], true ∈ bits, false ∈ bitsand(bits + false ' + bits3, bits4)

Rule argument

Additional information

Used simplifier rules: bitsand-onein: ⊦ ¬ true ∈ bits ∧ bits1 ≠ [] → ( true ∈ bitsand(bits + false ' + bits0, bits1) ↔ true ∈ bitsand(false ' + bits0, bits1));% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;