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Proof step 11 from proof disjoint-subset (previous step)

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, r0, r, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ ¬ junkref(str, r, st) ∧ st ⊆ st0 → (path∃(str, r, r0, st0) ↔ path∃(str, r, r0, st))
Proof step 10:
¬ (¬ junkref(str, r0, 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, r0, r, st)

Rule argument

[x0lemmaarg: spec-name: cyclic instance-name: lemma-name: pathex-subset-equiv antecedent: succedent: ¬ junkref(str, r, st) ∧ st ⊆ st0 → (path∃(str, r, r0, st0) ↔ path∃(str, r, r0, st)) substlist: [str, r, st, st0, r0] → [str, r0, st, st1, r] add preconditions?: F is rewrite lemma?: T current spec?: F rotate lemma?: F replace all?: T paths: ((1, 0, 2)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (str, r0, st, st1, r)]]