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Proof step 11 from proof lookup-list-put-ex (previous step)

∀ st0, r1, sk1, val0, r2. (st0[r1 -- sk1, val0][r2 -- sks0]) ≠ st0[r2 -- sks0] → (∃ sks, sks1. sks0 = sks + sk1 ' + sks1 ∧ st0[r2 -- sks] .refval = r1) ⊦ sks0 = [], (st[r0 -- sk, val][val .refval -- sks0]) = st[r0 -- (sk ' + sks0)], ∃ sks, sks1. sk ' + sks0 = sks + sk ' + sks1 ∧ st[r0 -- sks] .refval = r0

User applied rule exists right

And obtained the following new goals

Rule argument

[EXRARG : pos: right 3, [extquanttermlist: ([], sks0)abort:F, discard:T, computed:F]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, Nil, ⊦ [] + sks = sks], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : list, storekeys, app-01, ⊦ sks + sks0 = sks + sks1 ↔ sks0 = sks1], [CSIMPNAMEDSEQ : javavalue, , sel-09, ⊦ refval(r) .refval = r])], [EXTRATERMS : ([], sks0)]]