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Proof step 30 from proof and-insgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ bits = [], boolvar ∈ bitsand(sgn(bits), bits2 + boolvar0 ') ↔ (boolvar ↔ bits.last ∧ boolvar0) ∨ boolvar ∈ bitsand(bits.butlast, bits2), bits = bits.first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first '

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];