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Proof step 16 from proof unchanged-pathex (previous step)

path∃(str, r, r0, st), unchanged(str, r, st, st1) ⊦ ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st))
Proof step 15:
∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st), unchanged(str, r, st, st1) ⊦ ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st1)

Rule argument

[x0lemmaarg: spec-name: cyclic instance-name: lemma-name: pathex-def antecedent: succedent: path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs (str, refs, mkclasstype(`java.lang.Object' .class), st)) substlist: [st, r0, r, str] → [st, r0, r, str] add preconditions?: F is rewrite lemma?: T current spec?: F rotate lemma?: F replace all?: T paths: ((0, 0, 0)) dummy: () ]

Additional information

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