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Proof step 63 from proof and-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 64:
(¬ boolvar) ∈ bits2, ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits0 + true ' = bits0.first ', sgn(bits2 + boolvar ') = sgn(bitsand(bits0, bits2)) + boolvar ', bits2 = [], bits0 = [], bits2 + boolvar ' = bits2.first ', false ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);