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Proof step 15 from proof cyclic-subset (previous step)

st ⊆ st0, validrefs(str, r ' + refs0, st0[r -- _type] .type, st0) ⊦ cyclic(str, r, st), junkref(str, r, st), ¬ r ∈ refs0 ∧ ¬ dups(refs0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ cyclic(str, r, st) ↔ (∃ refs. dups(refs) ∧ refs .first = r ∧ validrefs(str, refs, st[r -- _type] .type, st))
Proof step 14:
st ⊆ st0, validrefs(str, r ' + refs0, st0[r -- _type] .type, st0) ⊦ ∃ refs. dups(refs) ∧ refs .first = r ∧ validrefs(str, refs, st[r -- _type] .type, st), junkref(str, r, st), ¬ r ∈ refs0 ∧ ¬ dups(refs0)

Rule argument

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

Additional information

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