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Proof step 22 from proof refin-del (previous step)

⊦ r = r1, r ∈ st -- r1 ↔ r ∈ st

User applied rule case distinction

And obtained the following new goals

Proof step 12:
r ∈ st -- r1 ⊦ r ∈ st, r = r1
Proof step 21:
r ∈ st ⊦ r ∈ st -- r1, r = r1

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]