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Proof step 21 from proof exists-fournew (previous step)

r < r0, jvmref < r, jvmref < r0, r0 < r1, r < r1, jvmref < r1, r1 < r2, r0 < r2, r < r2, jvmref < r2, r2 < r3, r1 < r3, r0 < r3, r < r3, jvmref < r3, ∀ r. ∃ r0. r < r0, ∀ r0. r0 ∈ st → r0 < r ⊦ r1 = r2, r0 = r2, r0 = r1, r = r2, r = r1, r = r0, r2 = jvmref, r = jvmref, r1 = jvmref, r0 = jvmref, (r2 -- sk ∈ st → r4 ≠ r2) ∧ (r1 -- sk1 ∈ st → r6 ≠ r1) ∧ (rk ∈ st → rk .ref ≠ r0) ∧ (r -- sk0 ∈ st → r5 ≠ r)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
rk = r7 -- sk2, r < r0, jvmref < r, jvmref < r0, r0 < r1, r < r1, jvmref < r1, r1 < r2, r0 < r2, r < r2, jvmref < r2, r2 < r3, r1 < r3, r0 < r3, r < r3, jvmref < r3, ∀ r. ∃ r0. r < r0, ∀ r0. r0 ∈ st → r0 < r ⊦ r1 = r2, r0 = r2, r0 = r1, r = r2, r = r1, r = r0, r2 = jvmref, r = jvmref, r1 = jvmref, r0 = jvmref, (r2 -- sk ∈ st → r4 ≠ r2) ∧ (r1 -- sk1 ∈ st → r6 ≠ r1) ∧ (rk ∈ st → r7 ≠ r0) ∧ (r -- sk0 ∈ st → r5 ≠ r)

Rule argument

[speclemmaarg: ,r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk ,[rk, r, sk] → [rk, r7, sk2] ,T ,refkey , ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk])], [EXTRATERMS : (rk, r7, sk2)]]