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Proof step 20 from proof pathex-def-simple (previous step)

⊦ path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ 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, mkclasstype(`java.lang.Object' .class), st))
Proof step 19:
⊦ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st)) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[refs .first -- _type] .type, st))

Rule argument

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

Additional information

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