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Proof step 8 from proof isarray-subset (previous step)

st[r -- _type] = typeval(mkarraytype(ty)), r ∈ st, st ⊆ st1, okarray(r, ty, st), r -- _type ∈ st, r ∈ st1, okarray(r, ty, st1) ⊦ r = jvmref, r -- _type ∈ st1 ∧ is_typevalue(st1[r -- _type]) ∧ is_arraytype(st1[r -- _type] .type) ∧ okarray(r, st1[r -- _type] .type .type, st1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk])
Proof step 7:
st[r -- _type] = typeval(mkarraytype(ty)), r ∈ st, ∀ rk. rk ∈ st → rk ∈ st1 ∧ st[rk] = st1[rk], okarray(r, ty, st), r -- _type ∈ st, r ∈ st1, okarray(r, ty, st1) ⊦ r = jvmref, r -- _type ∈ st1 ∧ is_typevalue(st1[r -- _type]) ∧ is_arraytype(st1[r -- _type] .type) ∧ okarray(r, st1[r -- _type] .type .type, st1)

Rule argument

[speclemmaarg: ,st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk]) ,[st2, st1] → [st1, st] ,F ,store ,javastore ,Subset ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st1, st)]]