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Proof step 32 from proof and-insgn (previous step)

bits = bits0 + boolvar1 ', ∀ 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 ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(bits0, bits2), bits = bits.first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ boolvar ∈ bitsand(sgn(bits0 + boolvar1 '), bits2 + boolvar0 ') ↔ (boolvar ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(bits0, bits2), bits0 + boolvar1 ' = (bits0 + boolvar1 ').first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first '

Rule argument

Additional information

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