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Proof step 21 from proof junkref-union-disjoint (previous step)

st1 .refs ∩ st .refs = ∅, junkref(str, r, st1 ∪ st), r ∈ st1 ⊦ junkref(str, r, st1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ junkref(str, r, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[r -- _type] .type, st))
Proof step 20:
st1 .refs ∩ st .refs = ∅, ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, (st1 ∪ st)[r -- _type] .type, st1 ∪ st), r ∈ st1 ⊦ junkref(str, r, st1)

Rule argument

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

Additional information

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