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Proof step 6 from proof lookup-put-ex-first (previous step)

⊦ val = st[r0 -- sk], ∃ sks, sks0. sk ' = sks + sk ' + sks0 ∧ st[r0 -- sks] .refval = r0 ∧ (∀ sks2. sks2 + sk ' ⊑ sks → st[r0 -- sks2] .refval ≠ st[r0 -- sks] .refval)

User applied rule exists right

And obtained the following new goals

Proof step 5:
⊦ st[r -- []] = refval(r)

Rule argument

[EXRARG : pos: right 2, [extquanttermlist: ([], [])abort:F, discard:T, computed:F]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, Nil, ⊦ [] + sks = sks], [CSIMPNAMEDSEQ : list, storekeys, app, ⊦ sks + [] = sks], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : list, storekeys, c-04, ⊦ sk ' = sk0 ' ↔ sk = sk0], [CSIMPNAMEDSEQ : javavalue, , sel-09, ⊦ refval(r) .refval = r], [CSIMPNAMEDSEQ : list-last, storekeys, eq-01-01, ⊦ sks ⊑ [] ↔ sks = []], [CSIMPNAMEDSEQ : list, storekeys, app-02-ant, ⊦ ¬ sks + sks0 ≠ [] ↔ sks = [] ∧ sks0 = []], [CSIMPNAMEDSEQ : list, storekeys, e, ⊦ sk ' ≠ []])], [EXTRATERMS : ([], [])]]