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Proof step 31 from proof lookup-put-ex-first (previous step)

∀ sks2. sks2 + sk ' ⊑ sks → st[st[r -- sk0] .refval -- sks2] .refval ≠ st[st[r -- sk0] .refval -- sks] .refval, r ≠ st[st[r -- sk] .refval -- sks] .refval ∨ sk0 ≠ sk ⊦ (st[st[st[r -- sk0] .refval -- sks] .refval -- sk, val][st[r -- sk0] .refval -- (sks + sk ' + sks1)]) = st[st[r -- sk0] .refval -- (sks + sk ' + sks1)], ∃ sks2, sks3. sk0 ' + sks + sk ' + sks1 = sks2 + sk ' + sks3 ∧ st[r -- sks2] .refval = st[st[r -- sk0] .refval -- sks] .refval ∧ (∀ sks3. sks3 + sk ' ⊑ sks2 → st[r -- sks3] .refval ≠ st[r -- sks2] .refval), st[r -- (sk0 ' + sks)] .refval = st[st[r -- sk0] .refval -- sks] .refval ∧ (sks0 + sk ' ⊑ sk0 ' + sks → st[r -- sks0] .refval ≠ st[r -- (sk0 ' + sks)] .refval)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ sks ≠ [] → st[r -- (sk ' + sks)] = st[st[r -- sk] .refval -- sks]
Proof step 16:
⊦ st[r -- []] = refval(r)
Proof step 17:
⊦ st[r -- sk '] = st[r -- sk]
Proof step 30:
sks ≠ [], ∀ sks2. sks2 + sk ' ⊑ sks → st[st[r -- sk0] .refval -- sks2] .refval ≠ st[st[r -- sk0] .refval -- sks] .refval, r ≠ st[st[r -- sk] .refval -- sks] .refval ∨ sk0 ≠ sk ⊦ (st[st[st[r -- sk0] .refval -- sks] .refval -- sk, val][st[r -- sk0] .refval -- (sks + sk ' + sks1)]) = st[st[r -- sk0] .refval -- (sks + sk ' + sks1)], ∃ sks2, sks3. sk0 ' + sks + sk ' + sks1 = sks2 + sk ' + sks3 ∧ st[r -- sks2] .refval = st[st[r -- sk0] .refval -- sks] .refval ∧ (∀ sks3. sks3 + sk ' ⊑ sks2 → st[r -- sks3] .refval ≠ st[r -- sks2] .refval), st[st[r -- sk0] .refval -- sks] .refval = st[st[r -- sk0] .refval -- sks] .refval ∧ (sks0 + sk ' ⊑ sk0 ' + sks → st[r -- sks0] .refval ≠ st[st[r -- sk0] .refval -- sks] .refval)

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: lookup-rec-01 antecedent: succedent: sks ≠ [] → st[r -- (sk ' + sks)] = st[st[r -- sk] .refval -- sks] substlist: [sks, sk, r, st] → [sks, sk0, r, st] add preconditions?: F is rewrite lemma?: T current spec?: T rotate lemma?: F replace all?: T paths: ((1, 0, 2, 1, 0, 1, 0, 1, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : javavalue, , sel-09, ⊦ refval(r) .refval = r], [CSIMPNAMEDSEQ : list, storekeys, Nil, ⊦ [] + sks = sks], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : list-last, storekeys, eq-01-01, ⊦ sks ⊑ [] ↔ sks = []], [CSIMPNAMEDSEQ : list, storekeys, e-02, ⊦ sks + sk ' ≠ []], [CSIMPNAMEDSEQ : list, storekeys, app, ⊦ sks + [] = sks], [CSIMPNAMEDSEQ : list-last, storekeys, ins-nil, ⊦ sks ⊑ sk ' ↔ ¬ ¬ (sks = [] ∨ sks = sk ')], [CSIMPNAMEDSEQ : list, storekeys, app-02-ant, ⊦ ¬ sks + sks0 ≠ [] ↔ sks = [] ∧ sks0 = []], [CSIMPNAMEDSEQ : list, storekeys, e, ⊦ sk ' ≠ []], [CSIMPNAMEDSEQ : list, storekeys, c-05, ⊦ sk ' = sks + sk0 ' ↔ sk = sk0 ∧ sks = []])], [EXTRATERMS : (sks, sk0, r, st)]]