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Proof step 3 from proof update-at-notin (previous step)

∀ r, st, r0, sk. (update(r, fis0, st)[r0 -- sk]) = st[r0 -- sk] ∨ sk ∈ fis0 ⊦ sk = sk0, (update(r, fis0, st)[r0 -- sk]) = st[r0 -- sk], sk ∈ fis0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [QUANTTERMLIST : (r, st, r0, sk)]]

Additional information

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