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Proof step 24 from proof lift-emp (previous step)

∀ 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) ⊦ mk(rs, f) = ∅ ∧ (r ∈ rs → g(r) = a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
∀ r0. ¬ r0 ∈ rs ⊦ mk(rs, f) = ∅ ∧ (r ∈ rs → g(r) = a)

Rule argument

Additional information

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