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Proof step 8 from proof instore-rec-01 (previous step)

sks0 ≠ [] → (∀ r, sk, st. r -- (sk ' + sks0) ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- sks0 ∈ st) ⊦ r -- (sk ' + sk0 ' + sks0) ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- (sk0 ' + sks0) ∈ st

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]