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Proof step 13 from proof instore-02 (previous step)

⊦ (∃ sk0. r -- sk0 ∈ st[r0 -- sk, val]) ↔ r = r0 ∨ (∃ sk. r -- sk ∈ st)

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∃ sk0. r -- sk0 ∈ st[r0 -- sk, val] ⊦ r = r0 ∨ (∃ sk. r -- sk ∈ st)
Proof step 12:
r = r0 ∨ (∃ sk. r -- sk ∈ st) ⊦ ∃ sk0. r -- sk0 ∈ st[r0 -- sk, val]

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]