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Proof step 7 from proof getbar-union (previous step)

¬ r ∈ st .refs ∨ ¬ r ∈ st0 .refs, ∀ r. ¬ r ∈ st .refs ∨ ¬ r ∈ st0 .refs, r ∈ st0, r ∈ st ⊦ getbytearray(r, st ∪ st0) = getbytearray(r, st0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ r ∈ st .refs ↔ r ∈ st
Proof step 6:
¬ r ∈ st ∨ ¬ r ∈ st0 .refs, ∀ r. ¬ r ∈ st .refs ∨ ¬ r ∈ st0 .refs, r ∈ st0, r ∈ st ⊦ getbytearray(r, st ∪ st0) = getbytearray(r, st0)

Rule argument

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

Additional information

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