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Proof step 109 from proof bitsor-bi-sgn (previous step)

bits3 = boolvar2 + bits4, bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3)) = bits2int(true ' + bits2), bitsor(boolvar1 ' + bits0, bits3) = true ' + bits2, false ∈ bits2, (¬ boolvar1) ∈ bits0, boolvar ∨ boolvar0, ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3) + true ') = bits2int(true ' + bits2) * 2 + 1, bits3 = [], ¬ boolvar ∧ ¬ boolvar0, (boolvar1 ∨ boolvar2) ∧ false ∈ bitsor(sgn(boolvar1 ' + bits0), bits3)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 110:
bits2int(bitsor(sgn(boolvar1 ' + bits0), boolvar2 ' + bits4)) = bits2int(true ' + bits2), bitsor(boolvar1 ' + bits0, boolvar2 ' + bits4) = true ' + bits2, false ∈ bits2, (¬ boolvar1) ∈ bits0, boolvar ∨ boolvar0, ∀ bits, bits1. # bits1 < #((boolvar2 + bits4) + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(bitsor(sgn(boolvar1 ' + bits0), boolvar2 ' + bits4) + true ') = bits2int(true ' + bits2) * 2 + 1, ¬ boolvar ∧ ¬ boolvar0, (boolvar1 ∨ boolvar2) ∧ false ∈ bitsor(sgn(boolvar1 ' + bits0), boolvar2 ' + bits4)

Rule argument

Additional information

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