Term Rewriting System R:

   [y, x]
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   mod(0, y) -> 0
   mod(s(x), 0) -> 0
   mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
   if_mod(true, x, y) -> mod(minus(x, y), y)
   if_mod(false, s(x), s(y)) -> s(x)

Innermost Termination of R to be shown.


R contains the following Dependency Pairs: 


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

Furthermore, R contains three SCCs.


SCC1:

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

Removing rules from R by ordering and analyzing Dependency Pairs, Usable Rules, and Usable Equations.

This is possible by using the following (C_E-compatible) Polynomial ordering.

Polynomial interpretation:
POL(LE(x_1, x_2)) = 1 + x_1 + x_2
POL(s(x_1)) = 1 + x_1

The following Dependency Pairs can be deleted:

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



 This transformation is resulting in no new subcycles.


SCC2:

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

Removing rules from R by ordering and analyzing Dependency Pairs, Usable Rules, and Usable Equations.

This is possible by using the following (C_E-compatible) Polynomial ordering.

Polynomial interpretation:
POL(s(x_1)) = 1 + x_1
POL(MINUS(x_1, x_2)) = 1 + x_1 + x_2

The following Dependency Pairs can be deleted:

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



 This transformation is resulting in no new subcycles.


SCC3:

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

On this Scc, a Narrowing SCC transformation can be performed.
 
As a result of transforming the rule 

   MOD(s(x), s(y)) -> IF_MOD(le(y, x), s(x), s(y))
three new Dependency Pairs
are created:


   MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))
   MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))
   MOD(s(0), s(s(x''))) -> IF_MOD(false, s(0), s(s(x'')))

The transformation is resulting in one subcycle:


SCC3.Nar1:

MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))
MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))
IF_MOD(true, x, y) -> MOD(minus(x, y), y)

On this Scc, a Narrowing SCC transformation can be performed.
 
As a result of transforming the rule 

   IF_MOD(true, x, y) -> MOD(minus(x, y), y)
two new Dependency Pairs
are created:


   IF_MOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))
   IF_MOD(true, x'', 0) -> MOD(x'', 0)

The transformation is resulting in one subcycle:


SCC3.Nar1.Nar1:

MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))
IF_MOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))
MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))

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

   IF_MOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))
two new Dependency Pairs
are created:


   IF_MOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(s(y''''), s(x'''')), s(s(x'''')))
   IF_MOD(true, s(x''''), s(0)) -> MOD(minus(x'''', 0), s(0))

The transformation is resulting in two subcycles:


SCC3.Nar1.Nar1.Inst1:

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

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

   IF_MOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(s(y''''), s(x'''')), s(s(x'''')))
one new Dependency Pair
is created:


   IF_MOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(y'''', x''''), s(s(x'''')))

The transformation is resulting in one subcycle:


SCC3.Nar1.Nar1.Inst1.Rewr1:

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

By using a polynomial ordering, at least one Dependency Pair of this SCC can be strictly oriented.

Additionally, the following rules can be oriented: 

   minus(s(x), s(y)) -> minus(x, y)
   minus(x, 0) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
POL(IF_MOD(x_1, x_2, x_3)) = x_2
POL(s(x_1)) = 1 + x_1
POL(le(x_1, x_2)) = 0
POL(MOD(x_1, x_2)) = 1 + x_1
POL(minus(x_1, x_2)) = x_1
POL(true) = 0
POL(0) = 0
POL(false) = 0

 resulting in no subcycles.


SCC3.Nar1.Nar1.Inst2:

IF_MOD(true, s(x''''), s(0)) -> MOD(minus(x'''', 0), s(0))
MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))

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

   IF_MOD(true, s(x''''), s(0)) -> MOD(minus(x'''', 0), s(0))
one new Dependency Pair
is created:


   IF_MOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))

The transformation is resulting in one subcycle:


SCC3.Nar1.Nar1.Inst2.Rewr1:

IF_MOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))
MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))

Removing rules from R by ordering and analyzing Dependency Pairs, Usable Rules, and Usable Equations.

This is possible by using the following (C_E-compatible) Polynomial ordering.

Polynomial interpretation:
POL(IF_MOD(x_1, x_2, x_3)) = x_1 + x_2 + x_3
POL(s(x_1)) = x_1
POL(MOD(x_1, x_2)) = x_1 + x_2
POL(true) = 0
POL(0) = 0

No Dependency Pairs can be deleted.

The following rules of R can be deleted: 

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


 This transformation is resulting in one new subcycle:


SCC3.Nar1.Nar1.Inst2.Rewr1.MRR1:

MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))
IF_MOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))

Removing rules from R by ordering and analyzing Dependency Pairs, Usable Rules, and Usable Equations.

This is possible by using the following (C_E-compatible) Polynomial ordering.

Polynomial interpretation:
POL(IF_MOD(x_1, x_2, x_3)) = x_1 + x_2 + x_3
POL(s(x_1)) = 1 + x_1
POL(MOD(x_1, x_2)) = x_1 + x_2
POL(true) = 0
POL(0) = 0

The following Dependency Pairs can be deleted:

   IF_MOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))



 This transformation is resulting in no new subcycles.




Innermost Termination of R successfully shown.


Duration: 11.54 seconds.