Term Rewriting System R:
[x, y, z, u, v]
f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v))
f(x, y, y) -> y
f(x, y, g(y)) -> x
f(x, x, y) -> x
f(g(x), x, y) -> y

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v))
F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation


Dependency Pairs:

F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v)
F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v))


Rules:


f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v))
f(x, y, y) -> y
f(x, y, g(y)) -> x
f(x, x, y) -> x
f(g(x), x, y) -> y


Strategy:

innermost




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

F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v)
one new Dependency Pair is created:

F(f(x, y, f(x'0, y''', z'')), u'', f(x, y, f(x''', y'''', v''))) -> F(f(x'0, y''', z''), u'', f(x''', y'''', v''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Argument Filtering and Ordering


Dependency Pairs:

F(f(x, y, f(x'0, y''', z'')), u'', f(x, y, f(x''', y'''', v''))) -> F(f(x'0, y''', z''), u'', f(x''', y'''', v''))
F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v))


Rules:


f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v))
f(x, y, y) -> y
f(x, y, g(y)) -> x
f(x, x, y) -> x
f(g(x), x, y) -> y


Strategy:

innermost




The following dependency pairs can be strictly oriented:

F(f(x, y, f(x'0, y''', z'')), u'', f(x, y, f(x''', y'''', v''))) -> F(f(x'0, y''', z''), u'', f(x''', y'''', v''))
F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v))


The following usable rules for innermost can be oriented:

f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v))
f(x, y, y) -> y
f(x, y, g(y)) -> x
f(x, x, y) -> x
f(g(x), x, y) -> y


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(g(x1))=  1 + x1  
  POL(F(x1, x2, x3))=  1 + x1 + x2 + x3  
  POL(f(x1, x2, x3))=  1 + x1 + x2 + x3  

resulting in one new DP problem.
Used Argument Filtering System:
F(x1, x2, x3) -> F(x1, x2, x3)
f(x1, x2, x3) -> f(x1, x2, x3)
g(x1) -> g(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳AFS
             ...
               →DP Problem 3
                 ↳Dependency Graph


Dependency Pair:


Rules:


f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v))
f(x, y, y) -> y
f(x, y, g(y)) -> x
f(x, x, y) -> x
f(g(x), x, y) -> y


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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