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Proof step 20 from proof pathex-subset-equiv (previous step)

st ⊆ st0 ⊦ path∃(str, r, r0, st0) ↔ path∃(str, r, r0, st), junkref(str, r, st)

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]