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Proof step 21 from proof lookup-list-put-ex (previous step)

r ≠ r0 ∨ sk0 ≠ sk, (st[r0 -- sk, val][st[r -- sk0] .refval -- sks0]) ≠ st[st[r -- sk0] .refval -- sks0] → (∃ sks, sks1. sks0 = sks + sk ' + sks1 ∧ st[st[r -- sk0] .refval -- sks] .refval = r0) ⊦ sks0 = [], (st[r0 -- sk, val][st[r -- sk0] .refval -- sks0]) = st[st[r -- sk0] .refval -- sks0], ∃ sks, sks1. sk0 ' + sks0 = sks + sk ' + sks1 ∧ st[r -- sks] .refval = r0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : list, storekeys, e-03, ⊦ sks + sk ' + sks0 ≠ []])]]