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Proof step 4 from proof add-in-02 (previous step)

r ∈ addobj(r0, class1, fis0, st), ∀ r, r0, class1, st. r ∈ addobj(r0, class1, fis0, st) ↔ r = r0 ∨ r ∈ st ⊦ r = r0, r ∈ st

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 2, [QUANTTERMLIST : (r, r0, class1, st)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, r0, class1, st)]]