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Proof step 16 from proof pseq-subset-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ ¬ cyclic(str, r, st) ∧ ¬ cyclic(str, r1, st1) ∧ ¬ junkref(str, r, st) ∧ ¬ junkref(str, r1, st1) → (ps=(r, st, r1, st1) ↔ ps=(r1, st1, r, st))
Proof step 15:
¬ cyclic(str, r, st) ∧ ¬ cyclic(str, r1, st1) ∧ ¬ junkref(str, r, st) ∧ ¬ junkref(str, r1, st1), st1 ⊆ st2, ps=(r1, st1, r, st) ⊦ junkref(str, r, st), cyclic(str, r, st), junkref(str, r1, st1), cyclic(str, r1, st1), ps=(r, st, r1, st2)

Rule argument

[crewrite-arg: pseq-sym , F , [str, r, st, r1, st1] → [str, r, st, r1, st1] ]

Additional information

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