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Proof step 5 from proof disjoint-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
r1 ∈ refs, r0 ∈ refs, path∃(str, r1, r, st1), path∃(str, r0, r, st1), st ⊆ st1, junkref(str, r1, st), ∀ r. r ∈ refs → ¬ junkref(str, r, st) ⊦ r = jvmref, r1 = r0, path∃(str, r1, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]