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Proof step 13 from proof refs-put (previous step)

⊦ r0 ∈ st[r -- sk, val] .refs ↔ r0 = r ∨ r0 ∈ st .refs

User applied rule case distinction

And obtained the following new goals

Proof step 7:
r0 ∈ st[r -- sk, val] .refs ⊦ r0 = r ∨ r0 ∈ st .refs
Proof step 12:
r0 = r ∨ r0 ∈ st .refs ⊦ r0 ∈ st[r -- sk, val] .refs

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]