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Theorem subseteq-union-diff-one
n < # asar, n0 < # asar ⊦ disj(asar, 0, # asar) → (asar[n] ⊆ (union(asar, 0, # asar) \ asar[n0]) ∪ as ↔ n ≠ n0 ∨ asar[n0] ⊆ as)
Type: A user defined theorem
Validity: The proof is valid.
Features: s, ls,
Comment:
Used by:
Used by units in proved state:
Proof exists?: Yes
View
the proof
Infos about the proof
Proved?: Yes
Interactions: 3
Proof steps: 8
Used lemmas:
disj-unique
,
in-index-in-union
,
Used spec theorems:
set-union
/address-list-set/
subdef
set-union
/address-list-set/
Difference
Used simplifier rules:
nat-basic
/
f
set-union
/address-list-set/
unin-01
set-union
/address-list-set/
unc
set-union
/address-list-set/
una
set-union
/address-list-set/
diff-union
set-basic
/address-list-set/
size-zero-01
set-basic
/address-list-set/
ne-01
nat-basic
/
lf-02