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Proof step 9 from proof subset-delete (previous step)

r ≠ rk .ref, st[rk] = st0[rk], rk ∈ st, rk ∈ st0 -- r, ∀ rk. rk ∈ st → rk ∈ st0 -- r ∧ st[rk] = (st0 -- r)[rk] ⊦ r ∈ st, rk ∈ st0 ∧ st[rk] = st0[rk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
st[rk] = ∅[rk], rk ∈ st, rk ∈ st0 -- r, ∀ rk. rk ∈ st → rk ∈ st0 -- r ∧ st[rk] = (st0 -- r)[rk] ⊦ r = rk .ref, r ∈ st, rk ∈ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]])]]