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Proof step 4 from proof haskeys-addobj (previous step)

∀ r, r0, sks, class, st. r = r0 ∨ (hasKeys(r, sks, addobj(r0, class, fis0, st)) ↔ hasKeys(r, sks, st)) ⊦ r = r0, hasKeys(r, sks, addobj(r0, class, fis0, st)) ↔ hasKeys(r, sks, st)

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (r, r0, sks, class, st) abort:F , discard:T , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, r0, sks, class, st)]]