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Proof step 91 from proof and-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 92:
(¬ boolvar0) ∈ bits0, (¬ boolvar0 ∨ ¬ boolvar) ∈ bitsand(bits0, bits2), ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits0 = [], bits2 = [], bits0 + boolvar0 ' = bits0.first ', bits2 + boolvar ' = bits2.first ', (¬ boolvar0 ∨ ¬ boolvar) ∈ bitsand(sgn(bits0), bits2)

Rule argument

Additional information

Used simplifier rules: and-notempty: ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ sgn(bits) = [] ↔ bits = [];% ⊦ bits = bits0 + bits ↔ bits0 = [];