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Proof step 5 from proof fill-first (previous step)

fillsgn(n, boolvar ' + bits0) = boolvar0 + bits ∧ fillsgn(n, boolvar ' + bits0) ≠ [] ⊦ boolvar0 ↔ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
fillsgn(n, boolvar ' + bits0) = boolvar0 ' + bits ⊦ boolvar0 ↔ boolvar

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 ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ boolvar + bits = boolvar ' + bits;