Termination Proof using AProVE

Term Rewriting System R:
[YS, X, XS, X1, X2, Y, L]
app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, n_{_app}(activate(XS), YS))
app(X1, X2) -> n_{_app}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, n_{_nil})), n_{_zWadr}(activate(XS), activate(YS)))
zWadr(X1, X2) -> n_{_zWadr}(X1, X2)
prefix(L) -> cons(nil, n_{_zWadr}(L, prefix(L)))
nil -> n_{_nil}
activate(n_{_app}(X1, X2)) -> app(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_nil}) -> nil
activate(n_{_zWadr}(X1, X2)) -> zWadr(X1, X2)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(cons(X, XS), YS) -> ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, n_{_nil}))
ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
PREFIX(L) -> NIL
PREFIX(L) -> PREFIX(L)
ACTIVATE(n_{_app}(X1, X2)) -> APP(X1, X2)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_nil}) -> NIL
ACTIVATE(n_{_zWadr}(X1, X2)) -> ZWADR(X1, X2)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, n_{_nil}))
ACTIVATE(n_{_zWadr}(X1, X2)) -> ZWADR(X1, X2)
ACTIVATE(n_{_app}(X1, X2)) -> APP(X1, X2)
APP(cons(X, XS), YS) -> ACTIVATE(XS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, n_{_app}(activate(XS), YS))
app(X1, X2) -> n_{_app}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, n_{_nil})), n_{_zWadr}(activate(XS), activate(YS)))
zWadr(X1, X2) -> n_{_zWadr}(X1, X2)
prefix(L) -> cons(nil, n_{_zWadr}(L, prefix(L)))
nil -> n_{_nil}
activate(n_{_app}(X1, X2)) -> app(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_nil}) -> nil
activate(n_{_zWadr}(X1, X2)) -> zWadr(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_zWadr}(X1, X2)) -> ZWADR(X1, X2)

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(cons(x₁, x₂)) = x₁ + x₂

POL(ZWADR(x₁, x₂)) = x₁ + x₂

POL(n__zWadr(x₁, x₂)) = 1 + x₁ + x₂

POL(n__nil) = 0

POL(n__app(x₁, x₂)) = x₁

POL(ACTIVATE(x₁)) = x₁

POL(APP(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, n_{_nil}))
ACTIVATE(n_{_app}(X1, X2)) -> APP(X1, X2)
APP(cons(X, XS), YS) -> ACTIVATE(XS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, n_{_app}(activate(XS), YS))
app(X1, X2) -> n_{_app}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, n_{_nil})), n_{_zWadr}(activate(XS), activate(YS)))
zWadr(X1, X2) -> n_{_zWadr}(X1, X2)
prefix(L) -> cons(nil, n_{_zWadr}(L, prefix(L)))
nil -> n_{_nil}
activate(n_{_app}(X1, X2)) -> app(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_nil}) -> nil
activate(n_{_zWadr}(X1, X2)) -> zWadr(X1, X2)
activate(X) -> X

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳DGraph

...

               →DP Problem 4

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

APP(cons(X, XS), YS) -> ACTIVATE(XS)
ACTIVATE(n_{_app}(X1, X2)) -> APP(X1, X2)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, n_{_app}(activate(XS), YS))
app(X1, X2) -> n_{_app}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, n_{_nil})), n_{_zWadr}(activate(XS), activate(YS)))
zWadr(X1, X2) -> n_{_zWadr}(X1, X2)
prefix(L) -> cons(nil, n_{_zWadr}(L, prefix(L)))
nil -> n_{_nil}
activate(n_{_app}(X1, X2)) -> app(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_nil}) -> nil
activate(n_{_zWadr}(X1, X2)) -> zWadr(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_app}(X1, X2)) -> APP(X1, X2)

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(cons(x₁, x₂)) = x₂

POL(n__app(x₁, x₂)) = 1 + x₁

POL(ACTIVATE(x₁)) = x₁

POL(APP(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳DGraph

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Remaining

Dependency Pair:

APP(cons(X, XS), YS) -> ACTIVATE(XS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, n_{_app}(activate(XS), YS))
app(X1, X2) -> n_{_app}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, n_{_nil})), n_{_zWadr}(activate(XS), activate(YS)))
zWadr(X1, X2) -> n_{_zWadr}(X1, X2)
prefix(L) -> cons(nil, n_{_zWadr}(L, prefix(L)))
nil -> n_{_nil}
activate(n_{_app}(X1, X2)) -> app(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_nil}) -> nil
activate(n_{_zWadr}(X1, X2)) -> zWadr(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, n_{_app}(activate(XS), YS))
app(X1, X2) -> n_{_app}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
zWadr(nil, YS) -> nil
zWadr(XS, nil) -> nil
zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, n_{_nil})), n_{_zWadr}(activate(XS), activate(YS)))
zWadr(X1, X2) -> n_{_zWadr}(X1, X2)
prefix(L) -> cons(nil, n_{_zWadr}(L, prefix(L)))
nil -> n_{_nil}
activate(n_{_app}(X1, X2)) -> app(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_nil}) -> nil
activate(n_{_zWadr}(X1, X2)) -> zWadr(X1, X2)
activate(X) -> X

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes

POL(cons(x₁, x₂))	= x₁ + x₂
POL(ZWADR(x₁, x₂))	= x₁ + x₂
POL(n__zWadr(x₁, x₂))	= 1 + x₁ + x₂
POL(n__nil)	= 0
POL(n__app(x₁, x₂))	= x₁
POL(ACTIVATE(x₁))	= x₁
POL(APP(x₁, x₂))	= x₁

POL(cons(x₁, x₂))	= x₂
POL(n__app(x₁, x₂))	= 1 + x₁
POL(ACTIVATE(x₁))	= x₁
POL(APP(x₁, x₂))	= x₁