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Proof step 5 from proof fill-sgn-03 (previous step)

fillsgn(n + 1, sgn(bits0)) = boolvar + bits ∧ fillsgn(n + 1, sgn(bits0)) ≠ [], # bits0 ≤ n ⊦ bits = fillsgn(n, bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
fillsgn(n + 1, sgn(bits0)) = boolvar ' + bits, # bits0 ≤ n ⊦ bits = fillsgn(n, bits0)

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;