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Proof step 5 from proof is_array-subset (previous step)

st ⊆ st1, type∃(str, st[r -- _type] .type), is_array(r, st), ¬ type∃(str, st1[r -- _type] .type) ∨ st1[r -- _type] .type .jtelemtype = void_type ∨ st1[r -- _type] .type .jtelemtype = abstract_type ⊦ st[r -- _type] .type .jtelemtype = void_type, st[r -- _type] .type .jtelemtype = abstract_type

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ rk ∈ st ∧ st ⊆ st1 → st1[rk] = st[rk]
Proof step 4:
⊦ ¬ r -- _type ∈ st → ¬ is_array(r, st)

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]], [CSIMPSEQ : ⊦ ∅[rk] = noval])], [EXTRATERMS : (r -- _type, st, st1)]]