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Proof step 3 from proof add-in-03 (previous step)

r -- sk0 ∈ st, r ∈ st, ∀ r, sk, st, r0, class1. r -- sk ∈ st ∧ r ∈ st → r -- sk ∈ addobj(r0, class1, fis0, st) ⊦ r -- sk0 ∈ addobj(r0, class1, fis0, st), r = r0 ∧ sk0 = sk

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 3, [QUANTTERMLIST : (r, sk0, st, r0, class1)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, sk0, st, r0, class1)]]