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Theorem prefixn-split-sublist
⊦ n0 + n1 = n ∧ n ≤ # x → prefixn(n, x) = prefixn(n0, x) + sublist(n0, n1, x)
Type: A user defined theorem
Validity: The proof is valid.
Features:
Comment:
Used by:
Used by units in proved state:
Proof exists?: Yes
View
the proof
Infos about the proof
Proved?: Yes
Interactions: 4
Proof steps: 10
Used lemmas:
prefixn
,
Used spec theorems:
list-del
/
first-append
list-del
/
sublist-append
list-del
/
sublist-restn-02
list-del
/
firstn-restn-eq
Used simplifier rules:
nat
/
fle-01
list-del
/
elim-firstn-restn
list
/
a
list
/
len
nat-basic
/
l-02
nat
/
lep
nat
/
le-def
nat-basic
/
lf-02
nat-basic
/
lf-01
nat-basic
/
f
nat-basic
/
ls-02
nat-basic
/
c
nat-basic
/
a
nat
/
ler
nat
/
lep-02