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Proof step 10 from proof cyclic-subset-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
st ⊆ st0, validrefs(str, r ' + refs0, st[r -- _type] .type, st) ⊦ junkref(str, r, st), ¬ r ∈ refs0 ∧ ¬ dups(refs0), (dups(refs0) ∨ r ∈ refs0) ∧ validrefs(str, r ' + refs0, st0[r -- _type] .type, st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (r ' + refs) .first = r], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ dups(r ' + refs) ↔ dups(refs) ∨ r ∈ refs])]]