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Proof step 10 from proof union-subset-left (previous step)

st .refs ∩ st0 .refs = ∅, r -- sk ∈ st, r ∈ st ⊦ st[r -- sk] = (st ∪ st0)[r -- sk]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ rk ∈ st0 → (st ∪ st0)[rk] = st[rk]
Proof step 9:
¬ ¬ r -- sk ∈ st0, st .refs ∩ st0 .refs = ∅, r -- sk ∈ st, r ∈ st ⊦ st[r -- sk] = (st ∪ st0)[r -- sk]

Rule argument

[crewrite-arg: union-at-left, F, [rk, st0, st] → [r -- sk, st0, st]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r -- sk, st0, st)]]