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Proof step 9 from proof lookup-list-put-ex (previous step)

¬ sks0 ≠ [], ∀ st0, r1, sk1, val0, r2. (st0[r1 -- sk1, val0][r2 -- sks0]) ≠ st0[r2 -- sks0] → (∃ sks, sks1. sks0 = sks + sk1 ' + sks1 ∧ st0[r2 -- sks] .refval = r1) ⊦ (st[r0 -- sk, val][r -- (sk0 ' + sks0)]) = st[r -- (sk0 ' + 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 3:
⊦ st[r -- sk '] = st[r -- sk]
Proof step 4:
⊦ st[r -- []] = refval(r)
Proof step 8:
⊦ (st[r0 -- sk, val][r -- sk0]) = st[r -- sk0], ∃ sks, sks0. sk0 ' = sks + sk ' + sks0 ∧ st[r -- sks] .refval = r0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : javavalue, , inj-03, ⊦ refval(r) = refval(r0) ↔ r = r0], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : list, storekeys, app, ⊦ sks + [] = sks])]]