Termination Proof using AProVE

Term Rewriting System R:
[X, Y, DX, DY, DDX]
din(der(plus(X, Y))) -> u21(din(der(X)), X, Y)
din(der(times(X, Y))) -> u31(din(der(X)), X, Y)
din(der(der(X))) -> u41(din(der(X)), X)
u21(dout(DX), X, Y) -> u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) -> dout(plus(DX, DY))
u31(dout(DX), X, Y) -> u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) -> dout(plus(times(X, DY), times(Y, DX)))
u41(dout(DX), X) -> u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) -> dout(DDX)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

DIN(der(plus(X, Y))) -> U21(din(der(X)), X, Y)
DIN(der(plus(X, Y))) -> DIN(der(X))
DIN(der(times(X, Y))) -> U31(din(der(X)), X, Y)
DIN(der(times(X, Y))) -> DIN(der(X))
DIN(der(der(X))) -> U41(din(der(X)), X)
DIN(der(der(X))) -> DIN(der(X))
U21(dout(DX), X, Y) -> U22(din(der(Y)), X, Y, DX)
U21(dout(DX), X, Y) -> DIN(der(Y))
U31(dout(DX), X, Y) -> U32(din(der(Y)), X, Y, DX)
U31(dout(DX), X, Y) -> DIN(der(Y))
U41(dout(DX), X) -> U42(din(der(DX)), X, DX)
U41(dout(DX), X) -> DIN(der(DX))

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

Dependency Pairs:

DIN(der(der(X))) -> DIN(der(X))
U41(dout(DX), X) -> DIN(der(DX))
DIN(der(der(X))) -> U41(din(der(X)), X)
DIN(der(times(X, Y))) -> DIN(der(X))
U31(dout(DX), X, Y) -> DIN(der(Y))
DIN(der(times(X, Y))) -> U31(din(der(X)), X, Y)
DIN(der(plus(X, Y))) -> DIN(der(X))
U21(dout(DX), X, Y) -> DIN(der(Y))
DIN(der(plus(X, Y))) -> U21(din(der(X)), X, Y)

Rules:

din(der(plus(X, Y))) -> u21(din(der(X)), X, Y)
din(der(times(X, Y))) -> u31(din(der(X)), X, Y)
din(der(der(X))) -> u41(din(der(X)), X)
u21(dout(DX), X, Y) -> u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) -> dout(plus(DX, DY))
u31(dout(DX), X, Y) -> u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) -> dout(plus(times(X, DY), times(Y, DX)))
u41(dout(DX), X) -> u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) -> dout(DDX)

Strategy:

innermost

The following dependency pair can be strictly oriented:

U41(dout(DX), X) -> DIN(der(DX))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

u31(dout(DX), X, Y) -> u32(din(der(Y)), X, Y, DX)
u21(dout(DX), X, Y) -> u22(din(der(Y)), X, Y, DX)
u41(dout(DX), X) -> u42(din(der(DX)), X, DX)
din(der(plus(X, Y))) -> u21(din(der(X)), X, Y)
din(der(times(X, Y))) -> u31(din(der(X)), X, Y)
din(der(der(X))) -> u41(din(der(X)), X)
u32(dout(DY), X, Y, DX) -> dout(plus(times(X, DY), times(Y, DX)))
u22(dout(DY), X, Y, DX) -> dout(plus(DX, DY))
u42(dout(DDX), X, DX) -> dout(DDX)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(plus(x₁, x₂)) = 0

POL(U31(x₁, x₂, x₃)) = 0

POL(u22(x₁, x₂, x₃, x₄)) = x₁

POL(dout(x₁)) = 1

POL(din(x₁)) = 0

POL(u21(x₁, x₂, x₃)) = 0

POL(u41(x₁, x₂)) = 0

POL(DIN(x₁)) = 0

POL(u32(x₁, x₂, x₃, x₄)) = x₁

POL(u42(x₁, x₂, x₃)) = x₁

POL(U41(x₁, x₂)) = x₁

POL(U21(x₁, x₂, x₃)) = 0

POL(times(x₁, x₂)) = 0

POL(u31(x₁, x₂, x₃)) = 0

POL(der(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Dependency Graph

Dependency Pairs:

DIN(der(der(X))) -> DIN(der(X))
DIN(der(der(X))) -> U41(din(der(X)), X)
DIN(der(times(X, Y))) -> DIN(der(X))
U31(dout(DX), X, Y) -> DIN(der(Y))
DIN(der(times(X, Y))) -> U31(din(der(X)), X, Y)
DIN(der(plus(X, Y))) -> DIN(der(X))
U21(dout(DX), X, Y) -> DIN(der(Y))
DIN(der(plus(X, Y))) -> U21(din(der(X)), X, Y)

Rules:

din(der(plus(X, Y))) -> u21(din(der(X)), X, Y)
din(der(times(X, Y))) -> u31(din(der(X)), X, Y)
din(der(der(X))) -> u41(din(der(X)), X)
u21(dout(DX), X, Y) -> u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) -> dout(plus(DX, DY))
u31(dout(DX), X, Y) -> u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) -> dout(plus(times(X, DY), times(Y, DX)))
u41(dout(DX), X) -> u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) -> dout(DDX)

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 3

                 ↳Polynomial Ordering

Dependency Pairs:

DIN(der(times(X, Y))) -> DIN(der(X))
U31(dout(DX), X, Y) -> DIN(der(Y))
DIN(der(times(X, Y))) -> U31(din(der(X)), X, Y)
DIN(der(plus(X, Y))) -> DIN(der(X))
U21(dout(DX), X, Y) -> DIN(der(Y))
DIN(der(plus(X, Y))) -> U21(din(der(X)), X, Y)
DIN(der(der(X))) -> DIN(der(X))

Rules:

din(der(plus(X, Y))) -> u21(din(der(X)), X, Y)
din(der(times(X, Y))) -> u31(din(der(X)), X, Y)
din(der(der(X))) -> u41(din(der(X)), X)
u21(dout(DX), X, Y) -> u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) -> dout(plus(DX, DY))
u31(dout(DX), X, Y) -> u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) -> dout(plus(times(X, DY), times(Y, DX)))
u41(dout(DX), X) -> u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) -> dout(DDX)

Strategy:

innermost

The following dependency pairs can be strictly oriented:

DIN(der(plus(X, Y))) -> DIN(der(X))
DIN(der(plus(X, Y))) -> U21(din(der(X)), X, Y)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(plus(x₁, x₂)) = 1 + x₁ + x₂

POL(U31(x₁, x₂, x₃)) = x₃

POL(u22(x₁, x₂, x₃, x₄)) = 0

POL(dout(x₁)) = 0

POL(din(x₁)) = 0

POL(u21(x₁, x₂, x₃)) = 0

POL(u41(x₁, x₂)) = 0

POL(DIN(x₁)) = x₁

POL(u32(x₁, x₂, x₃, x₄)) = 0

POL(u42(x₁, x₂, x₃)) = 0

POL(U21(x₁, x₂, x₃)) = x₃

POL(u31(x₁, x₂, x₃)) = 0

POL(times(x₁, x₂)) = x₁ + x₂

POL(der(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 4

                 ↳Dependency Graph

Dependency Pairs:

DIN(der(times(X, Y))) -> DIN(der(X))
U31(dout(DX), X, Y) -> DIN(der(Y))
DIN(der(times(X, Y))) -> U31(din(der(X)), X, Y)
U21(dout(DX), X, Y) -> DIN(der(Y))
DIN(der(der(X))) -> DIN(der(X))

Rules:

din(der(plus(X, Y))) -> u21(din(der(X)), X, Y)
din(der(times(X, Y))) -> u31(din(der(X)), X, Y)
din(der(der(X))) -> u41(din(der(X)), X)
u21(dout(DX), X, Y) -> u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) -> dout(plus(DX, DY))
u31(dout(DX), X, Y) -> u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) -> dout(plus(times(X, DY), times(Y, DX)))
u41(dout(DX), X) -> u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) -> dout(DDX)

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 5

                 ↳Polynomial Ordering

Dependency Pairs:

DIN(der(der(X))) -> DIN(der(X))
U31(dout(DX), X, Y) -> DIN(der(Y))
DIN(der(times(X, Y))) -> U31(din(der(X)), X, Y)
DIN(der(times(X, Y))) -> DIN(der(X))

Rules:

din(der(plus(X, Y))) -> u21(din(der(X)), X, Y)
din(der(times(X, Y))) -> u31(din(der(X)), X, Y)
din(der(der(X))) -> u41(din(der(X)), X)
u21(dout(DX), X, Y) -> u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) -> dout(plus(DX, DY))
u31(dout(DX), X, Y) -> u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) -> dout(plus(times(X, DY), times(Y, DX)))
u41(dout(DX), X) -> u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) -> dout(DDX)

Strategy:

innermost

The following dependency pairs can be strictly oriented:

DIN(der(times(X, Y))) -> U31(din(der(X)), X, Y)
DIN(der(times(X, Y))) -> DIN(der(X))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(plus(x₁, x₂)) = 0

POL(U31(x₁, x₂, x₃)) = x₃

POL(u22(x₁, x₂, x₃, x₄)) = 0

POL(dout(x₁)) = 0

POL(din(x₁)) = 0

POL(u21(x₁, x₂, x₃)) = 0

POL(u41(x₁, x₂)) = 0

POL(DIN(x₁)) = x₁

POL(u32(x₁, x₂, x₃, x₄)) = 0

POL(u42(x₁, x₂, x₃)) = 0

POL(u31(x₁, x₂, x₃)) = 0

POL(times(x₁, x₂)) = 1 + x₁ + x₂

POL(der(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pairs:

DIN(der(der(X))) -> DIN(der(X))
U31(dout(DX), X, Y) -> DIN(der(Y))

Rules:

din(der(plus(X, Y))) -> u21(din(der(X)), X, Y)
din(der(times(X, Y))) -> u31(din(der(X)), X, Y)
din(der(der(X))) -> u41(din(der(X)), X)
u21(dout(DX), X, Y) -> u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) -> dout(plus(DX, DY))
u31(dout(DX), X, Y) -> u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) -> dout(plus(times(X, DY), times(Y, DX)))
u41(dout(DX), X) -> u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) -> dout(DDX)

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pair:

DIN(der(der(X))) -> DIN(der(X))

Rules:

din(der(plus(X, Y))) -> u21(din(der(X)), X, Y)
din(der(times(X, Y))) -> u31(din(der(X)), X, Y)
din(der(der(X))) -> u41(din(der(X)), X)
u21(dout(DX), X, Y) -> u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) -> dout(plus(DX, DY))
u31(dout(DX), X, Y) -> u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) -> dout(plus(times(X, DY), times(Y, DX)))
u41(dout(DX), X) -> u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) -> dout(DDX)

Strategy:

innermost

The following dependency pair can be strictly oriented:

DIN(der(der(X))) -> DIN(der(X))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(DIN(x₁)) = 1 + x₁

POL(der(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

din(der(plus(X, Y))) -> u21(din(der(X)), X, Y)
din(der(times(X, Y))) -> u31(din(der(X)), X, Y)
din(der(der(X))) -> u41(din(der(X)), X)
u21(dout(DX), X, Y) -> u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) -> dout(plus(DX, DY))
u31(dout(DX), X, Y) -> u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) -> dout(plus(times(X, DY), times(Y, DX)))
u41(dout(DX), X) -> u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) -> dout(DDX)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(plus(x₁, x₂))	= 0
POL(U31(x₁, x₂, x₃))	= 0
POL(u22(x₁, x₂, x₃, x₄))	= x₁
POL(dout(x₁))	= 1
POL(din(x₁))	= 0
POL(u21(x₁, x₂, x₃))	= 0
POL(u41(x₁, x₂))	= 0
POL(DIN(x₁))	= 0
POL(u32(x₁, x₂, x₃, x₄))	= x₁
POL(u42(x₁, x₂, x₃))	= x₁
POL(U41(x₁, x₂))	= x₁
POL(U21(x₁, x₂, x₃))	= 0
POL(times(x₁, x₂))	= 0
POL(u31(x₁, x₂, x₃))	= 0
POL(der(x₁))	= 0

POL(plus(x₁, x₂))	= 1 + x₁ + x₂
POL(U31(x₁, x₂, x₃))	= x₃
POL(u22(x₁, x₂, x₃, x₄))	= 0
POL(dout(x₁))	= 0
POL(din(x₁))	= 0
POL(u21(x₁, x₂, x₃))	= 0
POL(u41(x₁, x₂))	= 0
POL(DIN(x₁))	= x₁
POL(u32(x₁, x₂, x₃, x₄))	= 0
POL(u42(x₁, x₂, x₃))	= 0
POL(U21(x₁, x₂, x₃))	= x₃
POL(u31(x₁, x₂, x₃))	= 0
POL(times(x₁, x₂))	= x₁ + x₂
POL(der(x₁))	= x₁