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Proof step 16 from proof fill-sgn-02 (previous step)

bits0 = boolvar + bits, fillsgn(# bits0, sgn(bits0)) = bits0 ⊦ fillsgn(1 + # bits0, sgn(bits0)) = boolvar ' + bits0, bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
fillsgn(1 + # bits, sgn(boolvar ' + bits)) = boolvar ' + bits ⊦ fillsgn(2 + # bits, sgn(boolvar ' + bits)) = boolvar ' + boolvar ' + bits

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ boolvar + bits = boolvar ' + bits;