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Proof step 3 from proof abs-of-emp (previous step)

⊦ ∀ fd. (fd ∈ abs-of(∅) ↔ fd ∈ ∅) ∧ abs-of(∅)[fd] = (∅)[fd]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ fd ∈ abs-of(of) ↔ fd ∈ of
Proof step 5:
⊦ abs-of(∅)[fd] = (∅)[fd]

Rule argument

Additional information

Used simplifier rules: abs-of-in: ⊦ fd ∈ abs-of(of) ↔ fd ∈ of;% ⊦ ¬ fd ∈ ∅; (s)% ⊦ ¬ fd ∈ ∅; (s)