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Proof step 5 from proof getbar-union-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
st .refs ∩ st0 .refs = ∅, r ∈ st, r ∈ st0 .refs ⊦ st0 .refs = ∅, getbytearray(r, i, j, st ∪ st0) = getbytearray(r, i, j, st0), # st0 .refs = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ rset ≠ ∅ → # rset ≠ 0], [CSIMPFORWARD : ⊦ r ∈ rset → rset ≠ ∅], [CSIMPSEQ : ⊦ rset ∩ rset0 = rset0 ∩ rset], [CSIMPSEQ : ⊦ (rset ∩ rset0) ∩ rset1 = rset ∩ rset0 ∩ rset1])]]