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Proof step 7 from proof lookup-put-other (conclusion)

⊦ ¬ (∃ sks0, sks1. sks = sks0 + (sk + sks1) ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks2. sks2 + sk ⊑ sks0 → st[r -- sks2] .refval ≠ r0)) → (st[r0 -- sk, val][r -- sks]) = st[r -- sks]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ (st[r0 -- sk, val][r -- sks]) = st[r -- sks], ∃ sks0, sks1. sks = sks0 + sk ' + sks1 ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks2. sks2 + sk ' ⊑ sks0 → st[r -- sks2] .refval ≠ st[r -- sks0] .refval)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, Last, ⊦ sks + sk = sks + sk '], [CSIMPNAMEDSEQ : list, storekeys, c, ⊦ sk + sks = sk ' + sks])]]