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Proof step 18 from proof cyclic-subset (previous step)

st ⊆ st0, ∃ refs. dups(refs) ∧ refs .first = r ∧ validrefs(str, refs, st0[r -- _type] .type, st0) ⊦ cyclic(str, r, st), junkref(str, r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
st ⊆ st0, dups(refs), validrefs(str, refs, st0[refs .first -- _type] .type, st0) ⊦ cyclic(str, refs .first, st), junkref(str, refs .first, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]