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Proof step 20 from proof and-insgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
∀ bits1, bits, boolvar. # bits1 < #(boolvar0 + bits2) → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ bits = true ', bits = false ', boolvar ∈ bitsand(sgn(bits), boolvar0 ' + bits2) ↔ boolvar ∈ bitsand(bits, boolvar0 ' + bits2), bits = [], boolvar0 ' + bits2 = true ', boolvar0 ' + bits2 = false ', bits2 ≠ [] ∧ bits ≠ bits.first '

Rule argument

Additional information

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