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Proof step 19 from proof lift-emp (previous step)

∀ r. ¬ r ∈ rs, ∀ r. r ∈ rs → g(r) = a ⊦ ∀ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
r ∈ rs, ∀ r. ¬ r ∈ rs, ∀ r. r ∈ rs → g(r) = a ⊦ # rs = 0, rs = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ rs ≠ ∅ → # rs ≠ 0; (forward) from specification set-basic[refset]%ne-01: ⊦ r ∈ rs → rs ≠ ∅; (forward) from specification set-basic[refset]% ⊦ ¬ r ∈ ∅; (s)% ⊦ r ∈ mk(rs, f, g) ↔ r ∈ rs; (s)