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Proof step 13 from proof disjoint-subset (previous step)

st ⊆ st1, path∃(str, r0, r, st1), path∃(str, r1, r, st1), r0 ∈ refs, r1 ∈ refs, ∀ r. r ∈ refs → ¬ junkref(str, r, st), ¬ path∃(str, r1, r, st) ∨ ¬ path∃(str, r0, r, st) ⊦ r1 = r0, r = jvmref

User applied rule case distinction

And obtained the following new goals

Proof step 7:
¬ path∃(str, r1, r, st), st ⊆ st1, path∃(str, r0, r, st1), path∃(str, r1, r, st1), r0 ∈ refs, r1 ∈ refs, ∀ r. r ∈ refs → ¬ junkref(str, r, st) ⊦ r1 = r0, r = jvmref
Proof step 12:
¬ path∃(str, r0, r, st), st ⊆ st1, path∃(str, r0, r, st1), path∃(str, r1, r, st1), r0 ∈ refs, r1 ∈ refs, ∀ r. r ∈ refs → ¬ junkref(str, r, st) ⊦ r1 = r0, r = jvmref

Rule argument

[CASEDARG : pos: left 7, ()]

Additional information

[[CONCRETESIMPRULES : ()]]