Termination Proof using AProVE

Term Rewriting System R:
[X, L, X1, X2]
a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_INCR}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_NATS} -> A_{_ADX}(a_{_zeros})
A_{_NATS} -> A_{_ZEROS}
A_{_HEAD}(cons(X, L)) -> MARK(X)
A_{_TAIL}(cons(X, L)) -> MARK(L)
MARK(incr(X)) -> A_{_INCR}(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(adx(X)) -> MARK(X)
MARK(nats) -> A_{_NATS}
MARK(zeros) -> A_{_ZEROS}
MARK(head(X)) -> A_{_HEAD}(mark(X))
MARK(head(X)) -> MARK(X)
MARK(tail(X)) -> A_{_TAIL}(mark(X))
MARK(tail(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, L)) -> MARK(L)
MARK(tail(X)) -> A_{_TAIL}(mark(X))
MARK(head(X)) -> MARK(X)
A_{_HEAD}(cons(X, L)) -> MARK(X)
MARK(head(X)) -> A_{_HEAD}(mark(X))
A_{_NATS} -> A_{_ADX}(a_{_zeros})
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(incr(X)) -> A_{_INCR}(mark(X))
A_{_INCR}(cons(X, L)) -> MARK(X)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

The following dependency pairs can be strictly oriented:

MARK(tail(X)) -> MARK(X)
MARK(tail(X)) -> A_{_TAIL}(mark(X))

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

a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(a__nats) = 0

POL(MARK(x₁)) = x₁

POL(adx(x₁)) = x₁

POL(A__NATS) = 0

POL(a__zeros) = 0

POL(A__ADX(x₁)) = x₁

POL(tail(x₁)) = 1 + x₁

POL(incr(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(A__TAIL(x₁)) = x₁

POL(a__adx(x₁)) = x₁

POL(A__INCR(x₁)) = x₁

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(nats) = 0

POL(nil) = 0

POL(a__tail(x₁)) = 1 + x₁

POL(s(x₁)) = x₁

POL(head(x₁)) = x₁

POL(zeros) = 0

POL(a__head(x₁)) = x₁

POL(A__HEAD(x₁)) = x₁

POL(a__incr(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Dependency Graph

Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
A_{_TAIL}(cons(X, L)) -> MARK(L)
MARK(head(X)) -> MARK(X)
A_{_HEAD}(cons(X, L)) -> MARK(X)
MARK(head(X)) -> A_{_HEAD}(mark(X))
A_{_NATS} -> A_{_ADX}(a_{_zeros})
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(incr(X)) -> A_{_INCR}(mark(X))
A_{_INCR}(cons(X, L)) -> MARK(X)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

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:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(head(X)) -> MARK(X)
A_{_HEAD}(cons(X, L)) -> MARK(X)
MARK(head(X)) -> A_{_HEAD}(mark(X))
A_{_NATS} -> A_{_ADX}(a_{_zeros})
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> A_{_INCR}(mark(X))
MARK(s(X)) -> MARK(X)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

The following dependency pairs can be strictly oriented:

MARK(head(X)) -> MARK(X)
MARK(head(X)) -> A_{_HEAD}(mark(X))

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

a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(a__nats) = 0

POL(MARK(x₁)) = x₁

POL(adx(x₁)) = x₁

POL(A__NATS) = 0

POL(a__zeros) = 0

POL(A__ADX(x₁)) = x₁

POL(tail(x₁)) = x₁

POL(incr(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(a__adx(x₁)) = x₁

POL(A__INCR(x₁)) = x₁

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(nats) = 0

POL(nil) = 0

POL(a__tail(x₁)) = x₁

POL(s(x₁)) = x₁

POL(head(x₁)) = 1 + x₁

POL(zeros) = 0

POL(a__head(x₁)) = 1 + x₁

POL(A__HEAD(x₁)) = x₁

POL(a__incr(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 4

                 ↳Dependency Graph

Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
A_{_HEAD}(cons(X, L)) -> MARK(X)
A_{_NATS} -> A_{_ADX}(a_{_zeros})
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> A_{_INCR}(mark(X))
MARK(s(X)) -> MARK(X)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

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:

MARK(s(X)) -> MARK(X)
A_{_NATS} -> A_{_ADX}(a_{_zeros})
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> A_{_INCR}(mark(X))
MARK(cons(X1, X2)) -> MARK(X1)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

The following dependency pair can be strictly oriented:

A_{_NATS} -> A_{_ADX}(a_{_zeros})

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

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(a__nats) = 1

POL(MARK(x₁)) = x₁

POL(adx(x₁)) = x₁

POL(A__NATS) = 1

POL(a__zeros) = 0

POL(A__ADX(x₁)) = x₁

POL(tail(x₁)) = x₁

POL(incr(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(a__adx(x₁)) = x₁

POL(A__INCR(x₁)) = x₁

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(nats) = 1

POL(nil) = 0

POL(a__tail(x₁)) = x₁

POL(s(x₁)) = x₁

POL(zeros) = 0

POL(head(x₁)) = x₁

POL(a__head(x₁)) = x₁

POL(a__incr(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> A_{_INCR}(mark(X))
MARK(cons(X1, X2)) -> MARK(X1)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

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 Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(adx(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> A_{_INCR}(mark(X))
MARK(s(X)) -> MARK(X)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

The following dependency pairs can be strictly oriented:

MARK(adx(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)

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

a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(a__nats) = 1

POL(MARK(x₁)) = x₁

POL(adx(x₁)) = 1 + x₁

POL(a__zeros) = 0

POL(A__ADX(x₁)) = 1 + x₁

POL(tail(x₁)) = x₁

POL(incr(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(a__adx(x₁)) = 1 + x₁

POL(A__INCR(x₁)) = x₁

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(nats) = 1

POL(nil) = 0

POL(a__tail(x₁)) = x₁

POL(s(x₁)) = x₁

POL(head(x₁)) = x₁

POL(zeros) = 0

POL(a__head(x₁)) = x₁

POL(a__incr(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> A_{_INCR}(mark(X))
MARK(s(X)) -> MARK(X)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

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

MARK(incr(X)) -> A_{_INCR}(mark(X))

10 new Dependency Pairs are created:

MARK(incr(incr(X''))) -> A_{_INCR}(a_{_incr}(mark(X'')))
MARK(incr(adx(X''))) -> A_{_INCR}(a_{_adx}(mark(X'')))
MARK(incr(nats)) -> A_{_INCR}(a_{_nats})
MARK(incr(zeros)) -> A_{_INCR}(a_{_zeros})
MARK(incr(head(X''))) -> A_{_INCR}(a_{_head}(mark(X'')))
MARK(incr(tail(X''))) -> A_{_INCR}(a_{_tail}(mark(X'')))
MARK(incr(nil)) -> A_{_INCR}(nil)
MARK(incr(cons(X1', X2'))) -> A_{_INCR}(cons(mark(X1'), X2'))
MARK(incr(s(X''))) -> A_{_INCR}(s(mark(X'')))
MARK(incr(0)) -> A_{_INCR}(0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 9

                 ↳Narrowing Transformation

Dependency Pairs:

MARK(incr(cons(X1', X2'))) -> A_{_INCR}(cons(mark(X1'), X2'))
MARK(incr(tail(X''))) -> A_{_INCR}(a_{_tail}(mark(X'')))
MARK(incr(head(X''))) -> A_{_INCR}(a_{_head}(mark(X'')))
MARK(incr(zeros)) -> A_{_INCR}(a_{_zeros})
MARK(incr(nats)) -> A_{_INCR}(a_{_nats})
MARK(incr(adx(X''))) -> A_{_INCR}(a_{_adx}(mark(X'')))
MARK(incr(incr(X''))) -> A_{_INCR}(a_{_incr}(mark(X'')))
MARK(s(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

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

MARK(adx(X)) -> A_{_ADX}(mark(X))

10 new Dependency Pairs are created:

MARK(adx(incr(X''))) -> A_{_ADX}(a_{_incr}(mark(X'')))
MARK(adx(adx(X''))) -> A_{_ADX}(a_{_adx}(mark(X'')))
MARK(adx(nats)) -> A_{_ADX}(a_{_nats})
MARK(adx(zeros)) -> A_{_ADX}(a_{_zeros})
MARK(adx(head(X''))) -> A_{_ADX}(a_{_head}(mark(X'')))
MARK(adx(tail(X''))) -> A_{_ADX}(a_{_tail}(mark(X'')))
MARK(adx(nil)) -> A_{_ADX}(nil)
MARK(adx(cons(X1', X2'))) -> A_{_ADX}(cons(mark(X1'), X2'))
MARK(adx(s(X''))) -> A_{_ADX}(s(mark(X'')))
MARK(adx(0)) -> A_{_ADX}(0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 10

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

MARK(adx(cons(X1', X2'))) -> A_{_ADX}(cons(mark(X1'), X2'))
MARK(adx(tail(X''))) -> A_{_ADX}(a_{_tail}(mark(X'')))
MARK(adx(head(X''))) -> A_{_ADX}(a_{_head}(mark(X'')))
MARK(adx(zeros)) -> A_{_ADX}(a_{_zeros})
MARK(adx(nats)) -> A_{_ADX}(a_{_nats})
MARK(adx(adx(X''))) -> A_{_ADX}(a_{_adx}(mark(X'')))
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(incr(X''))) -> A_{_ADX}(a_{_incr}(mark(X'')))
MARK(incr(tail(X''))) -> A_{_INCR}(a_{_tail}(mark(X'')))
MARK(incr(head(X''))) -> A_{_INCR}(a_{_head}(mark(X'')))
MARK(incr(zeros)) -> A_{_INCR}(a_{_zeros})
MARK(incr(nats)) -> A_{_INCR}(a_{_nats})
MARK(incr(adx(X''))) -> A_{_INCR}(a_{_adx}(mark(X'')))
MARK(incr(incr(X''))) -> A_{_INCR}(a_{_incr}(mark(X'')))
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(incr(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
MARK(incr(cons(X1', X2'))) -> A_{_INCR}(cons(mark(X1'), X2'))

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:12 minutes

POL(a__nats)	= 0
POL(MARK(x₁))	= x₁
POL(adx(x₁))	= x₁
POL(A__NATS)	= 0
POL(a__zeros)	= 0
POL(A__ADX(x₁))	= x₁
POL(tail(x₁))	= 1 + x₁
POL(incr(x₁))	= x₁
POL(mark(x₁))	= x₁
POL(A__TAIL(x₁))	= x₁
POL(a__adx(x₁))	= x₁
POL(A__INCR(x₁))	= x₁
POL(0)	= 0
POL(cons(x₁, x₂))	= x₁ + x₂
POL(nats)	= 0
POL(nil)	= 0
POL(a__tail(x₁))	= 1 + x₁
POL(s(x₁))	= x₁
POL(head(x₁))	= x₁
POL(zeros)	= 0
POL(a__head(x₁))	= x₁
POL(A__HEAD(x₁))	= x₁
POL(a__incr(x₁))	= x₁