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Theorem int2bytes-fillsgn-rec
⊦ 1 < # int2bytes(i) ∧ # int2bytes(i) ≤ n → fillsgn(n, int2bytes(i)) = fillsgn(n -1, int2bytes(i >> 8)) + i2b(i)
Type: A user defined theorem
Validity: The proof is valid.
Features:
Comment:
Used by:
setInt-theorem
,
setShort-theorem
,
Used by units in proved state:
Proof exists?: Yes
View
the proof
Infos about the proof
Proved?: Yes
Interactions: 14
Proof steps: 28
Used lemmas:
bytes2int-first-gez
,
bytes2int-pos
,
bytes2int-first-lez
,
bytes2int-neg
,
fillsgn-pos
,
int2int
,
fillsgn-neg
,
fillsgn-longer
,
int2bytes-byte
,
int2bytes-rec-shift
,
int2bytes-empty
,
Used spec theorems:
list
/bytelist/
first-01
Used simplifier rules:
nat
/
ler
nat
/
lep-02
nat-basic1
/
lf-01
nat
/
eqle-one
int-basic1
/
numlslsdel
nat-basic1
/
less-1
nat-basic2
/
eq-01
nat
/
cancel-01
nat-basic2
/
oz
nat
/
elim-pred-c
list-last
/bytelist/
mk-inj-01
list
/bytelist/
app-03
list
/bytelist/
e-02
list
/bytelist/
app-02-ant
int-abs
/
zero-gels-01
int-abs
/
forward-le-02
int-abs
/
forward-lsle
int-abs
/
zero-gels-04
intops
/
shiftright-not-neg
int-abs
/
forward-le
intops
/
shiftright-not-pos-01
int-basic1
/
numlsf
int-abs
/
forward-le-01
nat
/
ls-le-simp
nat
/
lels
nat-basic2
/
s
nat-basic1
/
neq-01
nat
/
ls
nat-basic2
/
succ-03
nat
/
les
nat
/
ls-03
nat
/
lels-01
nat-basic1
/
one-01
nat
/
ge
nat-basic1
/
f
nat
/
succ-01
nat-basic2
/
c
nat-basic2
/
a
nat-basic2
/
one
list
/bytelist/
a
list
/bytelist/
len
list
/bytelist/
len-02
int-abs
/
numleledel-01
int-abs
/
numlelsdel
int-abs
/
forward-lels
int-abs
/
lels-01
int-abs
/
lels
list-data
/bytelist/
l
nat-basic1
/
one
list
/bytelist/
Last
nat
/
pred-03