Term Rewriting System R:
[y, x, u, v, z]
merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if((x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MERGE(.(x, y), .(u, v)) -> IF((x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
MERGE(.(x, y), .(u, v)) -> MERGE(y, .(u, v))
MERGE(.(x, y), .(u, v)) -> MERGE(.(x, y), v)
++'(.(x, y), z) -> ++'(y, z)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`

Dependency Pairs:

MERGE(.(x, y), .(u, v)) -> MERGE(.(x, y), v)
MERGE(.(x, y), .(u, v)) -> MERGE(y, .(u, v))

Rules:

merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if((x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x

Strategy:

innermost

The following dependency pair can be strictly oriented:

MERGE(.(x, y), .(u, v)) -> MERGE(.(x, y), v)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(.(x1, x2)) =  1 + x2 POL(MERGE(x1, x2)) =  x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 3`
`             ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`

Dependency Pair:

MERGE(.(x, y), .(u, v)) -> MERGE(y, .(u, v))

Rules:

merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if((x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x

Strategy:

innermost

The following dependency pair can be strictly oriented:

MERGE(.(x, y), .(u, v)) -> MERGE(y, .(u, v))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(.(x1, x2)) =  1 + x2 POL(MERGE(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 3`
`             ↳Polo`
`             ...`
`               →DP Problem 4`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`

Dependency Pair:

Rules:

merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if((x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`

Dependency Pair:

++'(.(x, y), z) -> ++'(y, z)

Rules:

merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if((x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x

Strategy:

innermost

The following dependency pair can be strictly oriented:

++'(.(x, y), z) -> ++'(y, z)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(++'(x1, x2)) =  x1 POL(.(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if((x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes