Termination Proof using AProVE

Term Rewriting System R:
[X, L]
incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

INCR(cons(X, L)) -> INCR(L)
ADX(cons(X, L)) -> INCR(cons(X, adx(L)))
ADX(cons(X, L)) -> ADX(L)
NATS -> ADX(zeros)
NATS -> ZEROS
ZEROS -> ZEROS

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

Dependency Pair:

INCR(cons(X, L)) -> INCR(L)

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Strategy:

innermost

The following dependency pair can be strictly oriented:

INCR(cons(X, L)) -> INCR(L)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

INCR(x₁) -> INCR(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

Dependency Pair:

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
ZEROS -> ZEROS

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Strategy:
innermost
Dependency Pair:
ADX(cons(X, L)) -> ADX(L)

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
ZEROS -> ZEROS

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Strategy:
innermost
Dependency Pair:
ADX(cons(X, L)) -> ADX(L)

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Strategy:
innermost

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