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Proof step 26 from proof lookup-put-ex-first (previous step)

sks0 = sk1 + sks4, st[r -- sks0] .refval = st[st[r -- sk0] .refval -- sks] .refval, sks0 + sk ' ⊑ sk0 ' + sks, sks4 + sk ' ⊑ sks → st[st[r -- sk0] .refval -- sks4] .refval ≠ st[r -- sks0] .refval, r ≠ st[st[r -- sk] .refval -- sks] .refval ∨ sk0 ≠ sk ⊦ sks0 = [], (st[st[r -- sks0] .refval -- sk, val][st[r -- sk0] .refval -- (sks + sk ' + sks1)]) = st[st[r -- sk0] .refval -- (sks + sk ' + sks1)], sks = [], ∃ sks2, sks3. sk0 ' + sks + sk ' + sks1 = sks2 + sk ' + sks3 ∧ st[r -- sks2] .refval = st[r -- sks0] .refval ∧ (∀ sks3. sks3 + sk ' ⊑ sks2 → st[r -- sks3] .refval ≠ st[r -- sks2] .refval)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
st[r -- (sk0 ' + sks4)] .refval = st[st[r -- sk0] .refval -- sks] .refval, sks4 + sk ' ⊑ sks, r ≠ st[st[r -- sk] .refval -- sks] .refval ∨ sk0 ≠ sk ⊦ st[st[r -- sk0] .refval -- sks4] .refval = st[r -- (sk0 ' + sks4)] .refval, (st[st[r -- (sk0 ' + sks4)] .refval -- sk, val][st[r -- sk0] .refval -- (sks + sk ' + sks1)]) = st[st[r -- sk0] .refval -- (sks + sk ' + sks1)], sks = [], ∃ sks0, sks2. sk0 ' + sks + sk ' + sks1 = sks0 + sk ' + sks2 ∧ st[r -- sks0] .refval = st[r -- (sk0 ' + sks4)] .refval ∧ (∀ sks3. sks3 + sk ' ⊑ sks0 → st[r -- sks3] .refval ≠ st[r -- sks0] .refval)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list-last, storekeys, ins-01, ⊦ sk ' + sks ⊑ sk0 ' + sks0 ↔ sk = sk0 ∧ sks ⊑ sks0], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : list, storekeys, e-01, ⊦ sk ' + sks ≠ []], [CSIMPNAMEDSEQ : list, storekeys, c, ⊦ sk + sks = sk ' + sks])]]