Term Rewriting System R:
[x, y, z]
del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
=(nil, nil) -> true
=(.(x, y), nil) -> false
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))

Termination of R to be shown. 



   R
     ↳Overlay and local confluence Check



The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost. 


   R
     ↳OC
       →TRS2
         ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)
DEL(.(x, .(y, z))) -> ='(x, y)
F(true, x, y, z) -> DEL(.(y, z))
F(false, x, y, z) -> DEL(.(y, z))
='(.(x, y), .(u, v)) -> ='(x, u)
='(.(x, y), .(u, v)) -> ='(y, v)

Furthermore, R contains one SCC. 


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳Usable Rules (Innermost)


Dependency Pairs:

F(false, x, y, z) -> DEL(.(y, z))
F(true, x, y, z) -> DEL(.(y, z))
DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)


Rules:


del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
=(nil, nil) -> true
=(.(x, y), nil) -> false
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))


Strategy:

innermost




As we are in the innermost case, we can delete all 3 non-usable-rules.


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 2
                 ↳Negative Polynomial Order


Dependency Pairs:

F(false, x, y, z) -> DEL(.(y, z))
F(true, x, y, z) -> DEL(.(y, z))
DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)


Rules:


=(.(x, y), nil) -> false
=(nil, nil) -> true
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))


Strategy:

innermost




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

F(false, x, y, z) -> DEL(.(y, z))


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

=(.(x, y), nil) -> false
=(nil, nil) -> true
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))


Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, ..., x4) ) = x1 + x4 + 1

POL( false ) = 1

POL( DEL(x1) ) = x1

POL( .(x1, x2) ) = x2 + 1

POL( =(x1, x2) ) = 1

POL( true ) = 0

POL( and(x1, x2) ) = 0


This results in one new DP problem. 


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 3
                 ↳Negative Polynomial Order


Dependency Pairs:

F(true, x, y, z) -> DEL(.(y, z))
DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)


Rules:


=(.(x, y), nil) -> false
=(nil, nil) -> true
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))


Strategy:

innermost




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

F(true, x, y, z) -> DEL(.(y, z))


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

=(.(x, y), nil) -> false
=(nil, nil) -> true
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))


Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, ..., x4) ) = x1 + x4 + 1

POL( true ) = 1

POL( DEL(x1) ) = x1

POL( .(x1, x2) ) = x2 + 1

POL( =(x1, x2) ) = 1

POL( false ) = 0

POL( and(x1, x2) ) = 0


This results in one new DP problem. 


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pair:

DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)


Rules:


=(.(x, y), nil) -> false
=(nil, nil) -> true
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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