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Proof step 112 from proof bitsor-bi-sgn (previous step)

¬(¬ (boolvar1 ' + bits0).first ∧ ¬ (boolvar2 ' + bits4).first ∧ boolvar1 ' + bits0 ≠ [] ∧ boolvar2 ' + bits4 ≠ []), 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 113:
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), ¬ boolvar1 ∧ ¬ boolvar2

Rule argument

Additional information

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