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Proof step 44 from proof and-insgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', bits0 + true ' = (bits0 + true ').first ', (boolvar ↔ boolvar0) ∨ boolvar ∈ bits2 ↔ boolvar ∈ bitsand(bits0, bits2) ∨ (boolvar ↔ boolvar0), false ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;