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Proof step 26 from proof junkref-put-nopathex-equiv-01 (previous step)

is_valid_ref(str, r0, st) ⊦ sk = _type, junkref(str, r, st[r0 -- sk, val]) ↔ junkref(str, r, st), path∃(str, r, r0, st)

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]