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Proof step 5 from proof pseq-subset-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
ps=(r1, st1, r, st), st1 ⊆ st2, cyclic(str, r1, st2) ⊦ cyclic(str, r, st), cyclic(str, r1, st1), junkref(str, r, st), junkref(str, r1, st1), ps=(r, st, r1, st2), junkref(str, r1, st2)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]