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Proof step 5 from proof hasKeys-put-02 (previous step)

∀ r, st, sk, val. (hasKeys(r, sks0, st[r -- sk, val]) ↔ hasKeys(r, sks0, st)) ∨ sk ∈ sks0 ⊦ r -- sk0 ∈ st ∧ hasKeys(r, sks0, st[r -- sk, val]) ↔ r -- sk0 ∈ st ∧ hasKeys(r, sks0, st), sk = sk0, sk ∈ sks0

User applied rule all left

And closed the goal!

Rule argument

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

Additional information

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