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Proof step 43 from proof and-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits0 = [], bits2 + boolvar ' = (bits2 + boolvar ').first ', bits0 + false ' = bits0.first ', false ' = sgn(bitsand(bits0, bits2) + false '), true ∈ bits0

Rule argument

Additional information

Used simplifier rules: bitsand-zerol: ⊦ bitsand(false ', bits + boolvar ') = false ';% ⊦ sgn(boolvar ') = boolvar ';% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];