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Theorem exdivmod
j ≠ 0 ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j)
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: 13
Proof steps: 27
Used lemmas:
zero-left
,
one
,
dist-lsplit
,
minus-left
,
commutativity
,
associativity
,
zero-ge-add
,
exdivmodpos
,
Used spec theorems:
Used simplifier rules:
int-abs
/
abs-08
int-abs
/
sub-add-01
int-basic2
/
ls
int-abs
/
forward-lele
int-abs
/
forward-le-04
int-abs
/
zero-gels-04
int-abs
/
abs-neg
int-abs
/
neg
int-abs
/
neg-07
int-abs
/
sub-ls-01
int-abs
/
sub-eq
int-abs
/
Sub-def
int-abs
/
absle
int-abs
/
forward-le-02
int-abs
/
abs-16
int-basic1
/
lst-01
int-abs
/
forward-lsle
int-abs
/
forward-le-01
int-basic1
/
ls-01
int-abs
/
negle-07
int-abs
/
zero-gels-01
int-abs
/
lels-01
int-abs
/
abssub-02
int-abs
/
sub-add-09
int-abs
/
minus-ass
int-abs
/
neg-15
int-abs
/
subsub
int-abs
/
negeq
int-basic2
/
add-a
int-basic2
/
add-c
int-basic2
/
ls-02
int-abs
/
lels
int-abs
/
lep-01
int-abs
/
anti
int-abs
/
forward-le-05
int-basic2
/
add-zero
int-abs
/
elim-sub