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Theorem countodd-append-ind
m ≤ n, n ≤ n0, n0 ≤ # a ⊦ countodd(a, m, n, p) + countodd(a, n, n0, p) = countodd(a, m, n0, p)
Type: A user defined theorem
Validity: The proof is valid.
Features: ls,
Comment:
Used by:
countodd-maxcol0-bounded
,
Used by units in proved state:
Proof exists?: Yes
View
the proof
Infos about the proof
Proved?: Yes
Interactions: 3
Proof steps: 7
Used lemmas:
countodd-rec
,
countodd-base
,
Used spec theorems:
Used simplifier rules:
nat
/
let
nat
/
lez-01
nat-basic
/
a
nat-basic
/
c
nat
/
lez
list-data
/oarray-sort/
l
nat-basic
/
f
nat-basic
/
succ
nat
/
succ-01
nat
/
lels-03
nat
/
ler
nat-basic
/
eq-01
nat
/
le-def
nat
/
eqle-one
nat
/
le-one
nat
/
fle-01
nat-basic
/
lf-01
nat
/
fle
nat-basic
/
oz
nat
/
lep-02
nat
/
les
nat-basic
/
s-04
nat-basic
/
refl-nat
nat
/
le-ls-simp
nat
/
lels
nat
/
ls-le-simp
nat-basic
/
add-base