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Proof step 137 from proof bitsor-bi-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 138:
bitsor(boolvar1 ' + bits0, bits3) = false ' + bits2, bits2int([]) = bits2uint(bits2), bitsor(sgn(boolvar1 ' + bits0), bits3) = [], (¬ boolvar1) ∈ bits0, boolvar ∨ boolvar0, ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits3 = [], ¬ boolvar ∧ ¬ boolvar0

Rule argument

Additional information

Used simplifier rules: ⊦ bits2uint(bits) ≠ ~1;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ 2 * i = 0 ↔ i = 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i = i * (j + 1);% ⊦ i + 1 = ~1 ↔ i + 1 + 1 = 0;% ⊦ bits2int(true ') = ~1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;