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Proof step 38 from proof newrefs-three (previous step)

¬ newref(st[newref(st) -- 0 ', noval]) -- sk ∈ st[newref(st) -- 0 ', noval], newref(st[newref(st) -- 0 ', noval]) -- sk ∈ st, newref(st[newref(st) -- 0 ', noval]) ∈ st ⊦ ∃ r, r0, r1. newref_list(3, st) = r ' + r0 ' + r1 ' ∧ newref(r1, st) ∧ newref(r, st) ∧ newref(r0, st) ∧ r ≠ r0 ∧ r ≠ r1 ∧ r0 ≠ r1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 36:
⊦ rk ≠ rk0 → (rk ∈ st[rk0, val] ↔ rk ∈ st)
Proof step 37:
⊦ r ∈ st → newref(st) ≠ r

Rule argument

[speclemmaarg: ,rk ≠ rk0 → (rk ∈ st[rk0, val] ↔ rk ∈ st) ,[val, rk0, st, rk] → [noval, newref(st) -- 0 ', st, newref(st[newref(st) -- 0 ', noval]) -- sk] ,F ,store ,javastore ,in-neq ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0])], [EXTRATERMS : (noval, newref(st) -- 0 ', st, newref(st[newref(st) -- 0 ', noval]) -- sk)]]