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Proof step 3 from proof lookup-put-other (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
st[r -- sks] .refval = st[r -- sks1] .refval, sks + sk ' ⊑ sks1, ∀ sks. sks + sk ' ⊑ sks1 → st[r -- sks] .refval ≠ st[r -- sks1] .refval ⊦ (st[st[r -- sks1] .refval -- sk, val][r -- (sks1 + sk ' + sks0)]) = st[r -- (sks1 + sk ' + sks0)]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1])]]