Term Rewriting System R:
[x, y, z]
max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MAX(N(L(s(x)), L(s(y)))) -> MAX(N(L(x), L(y)))
MAX(N(L(x), N(y, z))) -> MAX(N(L(x), L(max(N(y, z)))))
MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

Furthermore, R contains two SCCs.

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

Dependency Pair:

MAX(N(L(s(x)), L(s(y)))) -> MAX(N(L(x), L(y)))

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MAX(N(L(s(x)), L(s(y)))) -> MAX(N(L(x), L(y)))

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(MAX(x1)) =  1 + x1 POL(L(x1)) =  x1 POL(N(x1, x2)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

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

Dependency Pair:

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

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:

MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(MAX(x1)) =  1 + x1 POL(L(x1)) =  0 POL(N(x1, x2)) =  1 + x2

resulting in one new DP problem.

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

Dependency Pair:

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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