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

st1 ⊆ st2, ps=(r1, st1, r, st) ⊦ cyclic(str, r, st), cyclic(str, r1, st2), junkref(str, r, st), junkref(str, r1, st2), ps=(r1, st2, r, st), junkref(str, r1, st1), cyclic(str, r1, st1)

User applied rule insert lemma

And obtained the following new goals

Proof step 11:
⊦ ¬ junkref(str, r, st) ∧ ¬ cyclic(str, r, st) ∧ ¬ junkref(str, r1, st1) ∧ ¬ cyclic(str, r1, st1) ∧ ps=(r, st, r1, st1) ∧ st ⊆ st2 → ps=(r, st2, r1, st1)

Rule argument

[LEMMAARG : pseq-subset, [str, r, st, r1, st1, st2] → [str, r1, st1, r, st, st2]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (str, r1, st1, r, st, st2)]]