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Proof step 15 from proof lift-prop (previous step)

r ∈ rs, ∀ r. (r ∈ mk(rs, f, g) ↔ r ∈ ∅) ∧ (r ∈ mk(rs, f, g) → mk(rs, f, g)[r]o = (∅)[r]o ∧ mk(rs, f, g)[r]a = (∅)[r]a) ⊦ # rs = 0, g(r) = a, rs = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
r ∈ rs, ∀ r0. ¬ r0 ∈ rs ⊦ # rs = 0, g(r) = a, rs = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ r ∈ ∅; (s)% ⊦ r ∈ mk(rs, f, g) ↔ r ∈ rs; (s)