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Proof step 12 from proof pathex-subset (previous step)

st ⊆ st0, path∃(str, r, r1, st), validref(str, r, ty, st) ⊦ path∃(str, r, r1, st0)

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 11:
st ⊆ st0, ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r1 ∧ validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st), validref(str, r, ty, st) ⊦ path∃(str, r, r1, st0)

Rule argument

[crewrite-arg: pathex-def, F, [st, r0, r, str] → [st, r1, r, str]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r1, r, str)]]