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Proof step 10 from proof lookup-subset (previous step)

st ⊆ st0, r -- sk ∈ st, st[r -- sk] .refval -- sks0 ∈ st, r ∈ st, ∀ r0, st1, st2. r0 -- sks0 ∈ st1 ∧ st1 ⊆ st2 → st2[r0 -- sks0] = st1[r0 -- sks0] ⊦ sks0 = [], st0[st0[r -- sk] .refval -- sks0] = st[st[r -- sk] .refval -- sks0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ rk ∈ st ∧ st ⊆ st1 → st1[rk] = st[rk]
Proof step 9:
r -- sk ∈ st ∧ st ⊆ st0, st ⊆ st0, r -- sk ∈ st, st[r -- sk] .refval -- sks0 ∈ st, r ∈ st, ∀ r0, st1, st2. r0 -- sks0 ∈ st1 ∧ st1 ⊆ st2 → st2[r0 -- sks0] = st1[r0 -- sks0] ⊦ sks0 = [], st0[st[r -- sk] .refval -- sks0] = st[st[r -- sk] .refval -- sks0]

Rule argument

[speclemmaarg: ,rk ∈ st ∧ st ⊆ st1 → st1[rk] = st[rk] ,[rk, st, st1] → [r -- sk, st, st0] ,F ,store ,javastore ,superset-elem-eq ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r -- sk, st, st0)]]