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Proof step 8 from proof sgn-in-01 (previous step)

bits0 = boolvar1 + bits, ∀ boolvar. boolvar ∈ sgn(bits0) ↔ boolvar ∈ bits0 ⊦ boolvar0 ∈ sgn(boolvar ' + bits0) ↔ (boolvar0 ↔ boolvar) ∨ boolvar0 ∈ bits0, bits0 = [], boolvar1 ↔ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ boolvar2. boolvar2 ∈ sgn(boolvar1 ' + bits) ↔ boolvar2 ∈ bits ∨ (boolvar2 ↔ boolvar1) ⊦ boolvar1 ↔ boolvar, boolvar0 ∈ sgn(boolvar ' + boolvar1 ' + bits) ↔ (boolvar0 ↔ boolvar) ∨ (boolvar0 ↔ boolvar1) ∨ boolvar0 ∈ bits

Rule argument

Additional information

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