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Proof step 9 from proof union-subset-left (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ rk ∈ st0 → (st ∪ st0)[rk] = st0[rk]
Proof step 8:
st .refs ∩ st0 .refs = ∅, r -- sk ∈ st0, r -- sk ∈ st, r ∈ st, r ∈ st0 ⊦ st[r -- sk] = st0[r -- sk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st])]]