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Proof step 23 from proof cyclic-def-shortest (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
r0 ∈ refs, validrefs(str, refs + r0 ', st[refs .first -- _type] .type, st) ⊦ dups(refs), ∃ refs0. dups(refs0) ∧ refs0 .first = refs .first ∧ validrefs(str, refs0, st[refs0 .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]