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Proof step 12 from proof cyclic-subset-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
st ⊆ st0, validrefs(str, r ' + refs0, st[r -- _type] .type, st) ⊦ junkref(str, r, st), ¬ r ∈ refs0 ∧ ¬ dups(refs0), ∃ refs. dups(refs) ∧ refs .first = r ∧ validrefs(str, refs, st0[refs .first -- _type] .type, st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]