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Proof step 40 from proof cyclic-put-nopathex (previous step)

⊦ sk = _type, cyclic(str, r, st[r0 -- sk, val]) ↔ cyclic(str, r, st), path∃(str, r, r0, st), junkref(str, r, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ cyclic(str, r, st) ↔ (∃ refs, r0. ¬ dups(refs) ∧ r0 ∈ refs ∧ refs .first = r ∧ validrefs(str, refs + r0 ', st[r -- _type] .type, st))
Proof step 39:
⊦ sk = _type, (∃ refs, r1. ¬ dups(refs) ∧ r1 ∈ refs ∧ refs .first = r ∧ validrefs(str, refs + r1 ', (st[r0 -- sk, val][r -- _type]) .type, st[r0 -- sk, val])) ↔ cyclic(str, r, st), path∃(str, r, r0, st), junkref(str, r, st)

Rule argument

[crewrite-arg: cyclic-def-shortest , F , [st, r, str] → [st[r0 -- sk, val], r, str] ]

Additional information

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