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Proof step 3 from proof in-put-keys (previous step)

∀ sk, sk0, r0, st, r, val. sk = sk0 ∨ (r0 -- sk0 ∈ st[r -- (sks0 + sk '), val] ↔ r0 -- sk0 ∈ st) ⊦ sk = sk0, r0 -- sk0 ∈ st[st[r -- sk1] .refval -- (sks0 + sk '), val] ↔ r0 -- sk0 ∈ st

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (sk, sk0, r0, st, st[r -- sk1] .refval, val) abort:F , discard:T , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk, sk0, r0, st, st[r -- sk1] .refval, val)]]