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Proof step 130 from proof bitsor-bi-sgn (previous step)

bits3 = boolvar1 + bits4, boolvar1, bitsor(false ' + bits0, bits3) = false ' + bits2, bits2int(bitsor(sgn(false ' + bits0), bits3)) = bits2uint(bits2), bitsor(sgn(false ' + bits0), bits3).first, true ∈ bits0, boolvar ∨ boolvar0, ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits3 = [], bits2int(bitsor(sgn(false ' + bits0), bits3) + true ') = 1 + 2 * bits2uint(bits2), ¬ boolvar ∧ ¬ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 131:
bitsor(false ' + bits0, true ' + bits4) = false ' + bits2, bits2int(bitsor(sgn(false ' + bits0), true ' + bits4)) = bits2uint(bits2), true ∈ bits0, bitsor(sgn(false ' + bits0), true ' + bits4).first, boolvar ∨ boolvar0, ∀ bits, bits1. # bits1 < #((true + bits4) + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(bitsor(sgn(false ' + bits0), true ' + bits4) + true ') = 1 + 2 * bits2uint(bits2), ¬ boolvar ∧ ¬ boolvar0

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;