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Proof step 96 from proof bitsor-bi-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 97:
bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3)) = ~1, bitsor(boolvar1 ' + bits0, bits3) = true ' + bits2, (¬ boolvar1) ∈ bits0, false ∈ bitsor(sgn(boolvar1 ' + bits0), bits3), boolvar ∨ boolvar0, ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits3 = [], bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3) + true ') = ~1, false ∈ bits2, ¬ boolvar ∧ ¬ boolvar0

Rule argument

Additional information

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