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Proof step 2 from proof junkref-union (previous step)

junkref(str, r, st0 ∪ st) ⊦ junkref(str, r, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ ¬ junkref(str, r, st) ∧ st ⊆ st1 → ¬ junkref(str, r, st1)

Rule argument

[crewrite-arg: junkref-subset , F , [str, r, st, st1] → [str, r, st, st0 ∪ st] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ st0 ⊆ st ∪ st0])], [EXTRATERMS : (str, r, st, st0 ∪ st)]]