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

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)) -> IF(le(x, s(y)), 0, p(minus(x, p(s(y)))))
MINUS(x, s(y)) -> LE(x, s(y))
MINUS(x, s(y)) -> P(minus(x, p(s(y))))
MINUS(x, s(y)) -> MINUS(x, p(s(y)))
MINUS(x, s(y)) -> P(s(y))

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


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


Strategy:

innermost




As we are in the innermost case, we can delete all 9 non-usable-rules.


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


Dependency Pair:

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


Rules:


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


Strategy:

innermost




As we are in the innermost case, we can delete all 8 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳Rewriting Transformation


Dependency Pair:

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


Rule:


p(s(x)) -> x


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

MINUS(x, s(y)) -> MINUS(x, p(s(y)))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 5
                 ↳Usable Rules (Innermost)


Dependency Pair:

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


Rule:


p(s(x)) -> x


Strategy:

innermost




As we are in the innermost case, we can delete all 1 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳Rw
             ...
               →DP Problem 6
                 ↳Size-Change Principle


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.

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