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Proof step 6 from proof subset-delete (previous step)

¬ 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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
st[rk] = (st0 -- rk .ref)[rk], rk ∈ st, rk ∈ st0 -- rk .ref, ∀ rk0. rk0 ∈ st → rk0 ∈ st0 -- rk .ref ∧ st[rk0] = (st0 -- rk .ref)[rk0] ⊦ rk .ref ∈ st, rk ∈ st0 ∧ st[rk] = st0[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]