Term Rewriting System R:
[x, y, z]
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))

Innermost Termination of R to be shown. 



   R
     ↳Removing Redundant Rules for Innermost Termination



Removing the following rules from R which left hand sides contain non normal subterms

div(div(x, y), z) -> div(x, times(y, z))


   R
     ↳RRRI
       →TRS2
         ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

PLUS(s(x), y) -> PLUS(x, y)
TIMES(s(x), y) -> PLUS(y, times(x, y))
TIMES(s(x), y) -> TIMES(x, y)
DIV(x, y) -> QUOT(x, y, y)
QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> DIV(x, s(z))

Furthermore, R contains three SCCs. 


   R
     ↳RRRI
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳Size-Change Principle
           →DP Problem 2
             ↳Neg POLO
           →DP Problem 3
             ↳SCP


Dependency Pair:

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


Rules:


plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))





We number the DPs as follows:
  1. PLUS(s(x), y) -> PLUS(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
     ↳RRRI
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳Negative Polynomial Order
           →DP Problem 3
             ↳SCP


Dependency Pairs:

QUOT(x, 0, s(z)) -> DIV(x, s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)
DIV(x, y) -> QUOT(x, y, y)


Rules:


plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))





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

QUOT(s(x), s(y), z) -> QUOT(x, y, z)


There are no usable rules (regarding the implicit AFS).
Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x1, ..., x3) ) = x1

POL( s(x1) ) = x1 + 1

POL( DIV(x1, x2) ) = x1


This results in one new DP problem. 


   R
     ↳RRRI
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳Neg POLO
             ...
               →DP Problem 4
                 ↳Instantiation Transformation
           →DP Problem 3
             ↳SCP


Dependency Pairs:

QUOT(x, 0, s(z)) -> DIV(x, s(z))
DIV(x, y) -> QUOT(x, y, y)


Rules:


plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))





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

DIV(x, y) -> QUOT(x, y, y)
one new Dependency Pair is created:

DIV(x'', s(z'')) -> QUOT(x'', s(z''), s(z''))

The transformation is resulting in no new DP problems. 



   R
     ↳RRRI
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳Neg POLO
           →DP Problem 3
             ↳Size-Change Principle


Dependency Pair:

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


Rules:


plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))





We number the DPs as follows:
  1. TIMES(s(x), y) -> TIMES(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