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Proof step 20 from proof newrefs-rec (previous step)

∀ rk. rk ∈ st → rk .ref ≠ r, ∀ r0. r0 ∈ refs → r0 ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r0), r0 ∈ refs ∨ r0 = r ⊦ r = jvmref, dups(refs), r ∈ refs, r0 ≠ jvmref ∧ (rk ∈ st → rk .ref ≠ r0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
rk = r1 -- sk, ∀ rk. rk ∈ st → rk .ref ≠ r, ∀ r0. r0 ∈ refs → r0 ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r0), r0 ∈ refs ∨ r0 = r ⊦ r = jvmref, dups(refs), r ∈ refs, r0 ≠ jvmref ∧ (rk ∈ st → r1 ≠ r0)

Rule argument

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

Additional information

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