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Proof step 45 from proof lookup-put-ex-first (conclusion)

(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 ≠ r0)

User applied rule structural induction

And obtained the following new goals

Proof step 1:
⊦ st[r -- []] = refval(r)
Proof step 44:
∀ st, r0, sk, val, r. (st[r0 -- sk, val][r -- sks0]) ≠ st[r -- sks0] → (∃ sks2, sks1. sks0 = sks2 + (sk + sks1) ∧ st[r -- sks2] .refval = r0 ∧ (∀ sks3. sks3 + sk ⊑ sks2 → st[r -- sks3] .refval ≠ r0)), (st[r0 -- sk, val][r -- (sk0 + sks0)]) ≠ st[r -- (sk0 + sks0)] ⊦ ∃ sks2, sks1. sk0 + sks0 = sks2 + (sk + sks1) ∧ st[r -- sks2] .refval = r0 ∧ (∀ sks3. sks3 + sk ⊑ sks2 → st[r -- sks3] .refval ≠ r0)

Rule argument

[VARWITHVARSEQSARG : sks, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : storekeys generated by [], +;], [CSIMPNAMEDSEQ : javavalue, , inj-03, ⊦ refval(r) = refval(r0) ↔ r = r0])]]