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Proof step 18 from proof disjoint-def-pathex (previous step)

(r2 ' + refs2) .last = (r3 ' + refs1) .last, validrefs(str, r3 ' + refs1, st[r3 -- _type] .type, st), validrefs(str, r2 ' + refs2, st[r2 -- _type] .type, st), r3 ∈ refs, r2 ∈ refs ⊦ (r3 ' + refs1) .last = jvmref, r2 = r3, path∃(str, r3, (r3 ' + refs1) .last, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[r -- _type] .type, st))
Proof step 17:
(r2 ' + refs2) .last = (r3 ' + refs1) .last, validrefs(str, r3 ' + refs1, st[r3 -- _type] .type, st), validrefs(str, r2 ' + refs2, st[r2 -- _type] .type, st), r3 ∈ refs, r2 ∈ refs ⊦ (r3 ' + refs1) .last = jvmref, r2 = r3, ∃ refs. refs ≠ [] ∧ refs .first = r3 ∧ refs .last = (r3 ' + refs1) .last ∧ validrefs(str, refs, st[r3 -- _type] .type, st)

Rule argument

[x0lemmaarg: spec-name: cyclic instance-name: lemma-name: pathex-def-simple antecedent: succedent: path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[r -- _type] .type, st)) substlist: [st, r0, r, str] → [st, (r3 ' + refs1) .last, r3, str] 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 : (st, (r3 ' + refs1) .last, r3, str)]]