Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(x, s(y)) -> PRED(minus(x, y))
MINUS(x, s(y)) -> MINUS(x, y)
MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) -> LE(y, x)
IFMOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
IFMOD(true, s(x), s(y)) -> MINUS(x, y)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Usable Rules (Innermost)
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules


Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)


Strategy:

innermost




As we are in the innermost case, we can delete all 11 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
           →DP Problem 4
Size-Change Principle
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules


Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. LE(s(x), s(y)) -> LE(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
Usable Rules (Innermost)
       →DP Problem 3
UsableRules


Dependency Pair:

MINUS(x, s(y)) -> MINUS(x, y)


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)


Strategy:

innermost




As we are in the innermost case, we can delete all 11 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
           →DP Problem 5
Size-Change Principle
       →DP Problem 3
UsableRules


Dependency Pair:

MINUS(x, s(y)) -> MINUS(x, y)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. MINUS(x, s(y)) -> MINUS(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1=1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
Usable Rules (Innermost)


Dependency Pairs:

IFMOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)


Strategy:

innermost




As we are in the innermost case, we can delete all 5 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 6
Negative Polynomial Order


Dependency Pairs:

IFMOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))


Rules:


minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false


Strategy:

innermost




The following Dependency Pair can be strictly oriented using the given order.

IFMOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false


Used ordering:
Polynomial Order with Interpretation:

POL( IFMOD(x1, ..., x3) ) = x2

POL( s(x1) ) = x1 + 1

POL( MOD(x1, x2) ) = x1

POL( minus(x1, x2) ) = x1

POL( pred(x1) ) = x1

POL( le(x1, x2) ) = 0

POL( true ) = 0

POL( false ) = 0


This results in one new DP problem.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 6
Neg POLO
             ...
               →DP Problem 7
Dependency Graph


Dependency Pair:

MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))


Rules:


minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes