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Proof step 17 from proof disjoint-subset (previous step)

disjoint(str, refs, 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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: disjoint-def-pathex antecedent: succedent: disjoint(str, refs, st) ↔ ¬ (∃ r, r0, r1. r ∈ refs ∧ r0 ∈ refs ∧ r ≠ r0 ∧ r1 ≠ jvmref ∧ path∃(str, r, r1, st) ∧ path∃(str, r0, r1, st)) substlist: [st, refs, str] → [st, refs, str] add preconditions?: F is rewrite lemma?: T current spec?: T rotate lemma?: F replace all?: T paths: ((0, 0, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, refs, str)]]