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Proof step 6 from proof getbar-union-02 (previous step)

st .refs ∩ st0 .refs = ∅, r ∈ st0, r ∈ st ⊦ getbytearray(r, i, j, st ∪ st0) = getbytearray(r, i, j, st0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ r ∈ st .refs ↔ r ∈ st
Proof step 5:
st .refs ∩ st0 .refs = ∅, r ∈ st0 .refs, r ∈ st ⊦ getbytearray(r, i, j, st ∪ st0) = getbytearray(r, i, j, st0)

Rule argument

[speclemmaarg: ,r ∈ st .refs ↔ r ∈ st ,[st, r] → [st0, r] ,T ,storefunsx , ,refs-def ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0, r)]]