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Proof step 12 from proof sgn-in-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
∀ boolvar. boolvar ∈ sgn(boolvar1 ' + bits) ↔ boolvar ∈ bits ∨ (boolvar ↔ boolvar1) ⊦ boolvar0 ∈ sgn(boolvar1 ' + bits) ↔ boolvar0 ∈ bits ∨ (boolvar0 ↔ boolvar1)

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) → (boolvar ∈ boolvar0 ' + bits ↔ boolvar ∈ bits);% ⊦ boolvar + bits = boolvar ' + bits;