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Proof step 10 from proof sgn-ex (previous step)

boolvar ' + bits ≠ [] ∧ ((boolvar ' + bits).first ↔ boolvar0), ∀ boolvar. ∃ n. boolvar ' + bits = mklist(boolvar, n) + sgn(boolvar ' + bits) ⊦ ∃ n. boolvar0 ' + boolvar ' + bits = mklist(boolvar0, n) + sgn(boolvar ' + bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ (boolvar ' + bits).first ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];