Termination Proof using AProVE

Term Rewriting System R:
[z, x, y]
app(app(h, z), app(e, x)) -> app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) -> app(e, 0)
app(app(d, z), app(app(g, x), y)) -> app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) -> app(e, app(app(g, x), y))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(h, z), app(e, x)) -> APP(app(h, app(c, z)), app(app(d, z), x))
APP(app(h, z), app(e, x)) -> APP(h, app(c, z))
APP(app(h, z), app(e, x)) -> APP(c, z)
APP(app(h, z), app(e, x)) -> APP(app(d, z), x)
APP(app(h, z), app(e, x)) -> APP(d, z)
APP(app(d, z), app(app(g, 0), 0)) -> APP(e, 0)
APP(app(d, z), app(app(g, x), y)) -> APP(app(g, app(e, x)), app(app(d, z), y))
APP(app(d, z), app(app(g, x), y)) -> APP(g, app(e, x))
APP(app(d, z), app(app(g, x), y)) -> APP(e, x)
APP(app(d, z), app(app(g, x), y)) -> APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> APP(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> APP(g, app(app(d, app(c, z)), app(app(g, x), y)))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> APP(app(d, z), app(app(g, x), y))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> APP(d, z)
APP(app(g, app(e, x)), app(e, y)) -> APP(e, app(app(g, x), y))
APP(app(g, app(e, x)), app(e, y)) -> APP(app(g, x), y)
APP(app(g, app(e, x)), app(e, y)) -> APP(g, x)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Size-Change Principle

       →DP Problem 2

         ↳MRR

       →DP Problem 3

         ↳Remaining

Dependency Pair:

APP(app(g, app(e, x)), app(e, y)) -> APP(app(g, x), y)

Rules:

app(app(h, z), app(e, x)) -> app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) -> app(e, 0)
app(app(d, z), app(app(g, x), y)) -> app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) -> app(e, app(app(g, x), y))

The original DP problem is in applicative form. Its DPs and usable rules are the following.

APP(app(g, app(e, x)), app(e, y)) -> APP(app(g, x), y)

none

It is proper and hence, it can be A-transformed which results in the DP problem

G(e(x), e(y)) -> G(x, y)

none

We number the DPs as follows:

G(e(x), e(y)) -> G(x, y)

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1
2	>	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

e(x₁) -> e(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳SCP

       →DP Problem 2

         ↳Modular Removal of Rules

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> APP(app(d, z), app(app(g, x), y))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(d, z), app(app(g, x), y)) -> APP(app(d, z), y)

Rules:

app(app(h, z), app(e, x)) -> app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) -> app(e, 0)
app(app(d, z), app(app(g, x), y)) -> app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) -> app(e, app(app(g, x), y))

The original DP problem is in applicative form. Its DPs and usable rules are the following.

app(app(g, app(e, x)), app(e, y)) -> app(e, app(app(g, x), y))

It is proper and hence, it can be A-transformed which results in the DP problem

D(c(z), g(g(x, y), 0)) -> D(z, g(x, y))
D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))
D(z, g(x, y)) -> D(z, y)

g(e(x), e(y)) -> e(g(x, y))

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(c(x₁)) = 1 + x₁

POL(0) = 0

POL(g(x₁, x₂)) = x₁ + x₂

POL(e(x₁)) = x₁

POL(D(x₁, x₂)) = 1 + x₁ + x₂

We have the following set D of usable symbols: {c, g, e, D}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

D(c(z), g(g(x, y), 0)) -> D(z, g(x, y))
D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))

4 non usable rules have been deleted.

The result of this processor delivers one new DP problem.
Note that we keep the A-transformed DP problem as result of this processor.

     ↳DPs

       →DP Problem 1

         ↳SCP

       →DP Problem 2

         ↳MRR

           →DP Problem 4

             ↳Modular Removal of Rules

       →DP Problem 3

         ↳Remaining

Dependency Pair:

D(z, g(x, y)) -> D(z, y)

Rule:

g(e(x), e(y)) -> e(g(x, y))

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(g(x₁, x₂)) = x₁ + x₂

POL(D(x₁, x₂)) = x₁ + x₂

We have the following set D of usable symbols: {D}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

D(z, g(x, y)) -> D(z, y)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

     ↳DPs

       →DP Problem 1

         ↳SCP

       →DP Problem 2

         ↳MRR

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

APP(app(h, z), app(e, x)) -> APP(app(h, app(c, z)), app(app(d, z), x))

Rules:

app(app(h, z), app(e, x)) -> app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) -> app(e, 0)
app(app(d, z), app(app(g, x), y)) -> app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) -> app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) -> app(e, app(app(g, x), y))

The Proof could not be continued due to a Timeout.
Termination of R could not be shown.
Duration:
1:00 minutes

POL(c(x₁))	= 1 + x₁
POL(0)	= 0
POL(g(x₁, x₂))	= x₁ + x₂
POL(e(x₁))	= x₁
POL(D(x₁, x₂))	= 1 + x₁ + x₂