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Proof step 25 from proof refs-equal (previous step)

# st .refs = # st0 .refs, r0 ∈ st0 .refs, r0 ∈ st .refs, r ∈ st0, rk ∈ st -- r0, rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ r = r0, st0 = ∅, st = ∅, r0 = rk .ref, r ∈ st, rk ∈ st0 -- r0 ∧ st[rk] = st0[rk]

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

[speclemmaarg: ,r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk ,[rk, r, sk] → [rk, r1, sk] ,T ,refkey , ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st], [CSIMPELIM : ⊦ r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk])], [EXTRATERMS : (rk, r1, sk)]]