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Proof step 27 from proof cyclic-def-shortest (conclusion)

⊦ cyclic(str, r, st) ↔ (∃ refs, r0. ¬ dups(refs) ∧ r0 ∈ refs ∧ refs .first = r ∧ validrefs(str, refs + r0 ', st[r -- _type] .type, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
⊦ cyclic(str, r, st) ↔ (∃ refs, r0. ¬ dups(refs) ∧ r0 ∈ refs ∧ refs .first = r ∧ validrefs(str, refs + r0 ', st[refs .first -- _type] .type, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]