Term Rewriting System R:
[x, y, z]
app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

APP(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(app(., app(app(., x), y)), z) -> APP(., y)
APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))
APP(i, app(app(., x), y)) -> APP(., app(i, y))
APP(i, app(app(., x), y)) -> APP(i, y)
APP(i, app(app(., x), y)) -> APP(i, x)

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`

Dependency Pairs:

APP(i, app(app(., x), y)) -> APP(i, x)
APP(i, app(app(., x), y)) -> APP(i, y)
APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
APP(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

The following dependency pairs can be strictly oriented:

APP(i, app(app(., x), y)) -> APP(i, x)
APP(i, app(app(., x), y)) -> APP(i, y)
APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)

Additionally, the following rules can be oriented:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(i) =  0 POL(1) =  0 POL(.) =  1 POL(app(x1, x2)) =  x1 + x2 POL(APP(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 2`
`             ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))
APP(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))

Rules:

app(app(., 1), x) -> x
app(app(., x), 1) -> x
app(app(., app(i, x)), x) -> 1
app(app(., x), app(i, x)) -> 1
app(app(., app(i, y)), app(app(., y), z)) -> z
app(app(., y), app(app(., app(i, y)), z)) -> z
app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
app(i, 1) -> 1
app(i, app(i, x)) -> x
app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))

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