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Proof step 10 from proof subset-delete (previous step)

st[rk] = (st0 -- r)[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]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ r ≠ rk .ref → (st -- r)[rk] = st[rk]
Proof step 6:
¬ r ≠ rk .ref, st[rk] = (st0 -- r)[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]
Proof step 9:
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]

Rule argument

[crewrite-arg: delete-at, F, [rk, r, st] → [rk, r, st0]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk, r, st0)]]