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Proof step 6 from proof disjoint-nopath-both (previous step)

disjoint(str, r ' + r0 ', st), path∃(str, r, r2, st), path∃(str, r0, r2, st), is_valid_ref(str, r2, st) ⊦ r = r0, r2 = jvmref

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 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))
Proof step 5:
∀ r3, r2, r1. ¬ (r3 ∈ r ' + r0 ' ∧ r2 ∈ r ' + r0 ' ∧ r3 ≠ r2 ∧ r1 ≠ jvmref ∧ path∃(str, r3, r1, st) ∧ path∃(str, r2, r1, st)), path∃(str, r, r2, st), path∃(str, r0, r2, st), is_valid_ref(str, r2, st) ⊦ r = r0, r2 = jvmref

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, r ' + r0 ', 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, r ' + r0 ', str)]]