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Proof step 5 from proof update-put-notin (previous step)

∀ r0, st, sk, val. (update(r0, fis0, st)[r0 -- sk, val]) = update(r0, fis0, st[r0 -- sk, val]) ∨ sk ∈ fis0 ⊦ (update(r0, fis0, st)[r0 -- sk, val][r0 -- sk0, val0]) = (update(r0, fis0, st[r0 -- sk, val])[r0 -- sk0, val0]), sk = sk0, sk ∈ fis0

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (r0, st, sk, val) abort:F , discard:T , computed:F ]]

Additional information

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