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Proof step 27 from proof pathex-def-nodups (previous step)

⊦ r = r0, path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ ¬ dups(refs) ∧ validrefs(str, refs, st[refs .first -- _type] .type, st))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[r -- _type] .type, st))
Proof step 26:
⊦ r = r0, (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[r -- _type] .type, st)) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ ¬ dups(refs) ∧ validrefs(str, refs, st[refs .first -- _type] .type, st))

Rule argument

[crewrite-arg: pathex-def-simple , F , [st, r0, r, str] → [st, r0, r, str] ]

Additional information

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