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Proof step 11 from proof instore-02 (previous step)

r -- sk0 ∈ st ∨ r = r0 ⊦ ∃ sk1. r = r0 ∧ sk1 = sk ∨ r -- sk1 ∈ st

User applied rule case distinction

And obtained the following new goals

Proof step 8:
r -- sk0 ∈ st ⊦ ∃ sk1. r = r0 ∧ sk1 = sk ∨ r -- sk1 ∈ st
Proof step 10:
r = r0 ⊦ ∃ sk1. r = r0 ∧ sk1 = sk ∨ r -- sk1 ∈ st

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]