Term Rewriting System R:
[y, x, z]
f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Innermost Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

F(.(nil, y)) -> F(y)
F(.(.(x, y), z)) -> F(.(x, .(y, z)))
G(.(x, nil)) -> G(x)
G(.(x, .(y, z))) -> G(.(.(x, y), z))

Furthermore, R contains two SCCs.

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

Dependency Pairs:

F(.(.(x, y), z)) -> F(.(x, .(y, z)))
F(.(nil, y)) -> F(y)

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(.(nil, y)) -> F(y)

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

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

resulting in one new DP problem.

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

Dependency Pair:

F(.(.(x, y), z)) -> F(.(x, .(y, z)))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(.(.(x, y), z)) -> F(.(x, .(y, z)))

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

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

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, 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 Pairs:

G(.(x, .(y, z))) -> G(.(.(x, y), z))
G(.(x, nil)) -> G(x)

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

G(.(x, nil)) -> G(x)

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

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

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 5
Polynomial Ordering

Dependency Pair:

G(.(x, .(y, z))) -> G(.(.(x, y), z))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

G(.(x, .(y, z))) -> G(.(.(x, y), z))

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(G(x1)) =  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
Polo
...
→DP Problem 6
Dependency Graph

Dependency Pair:

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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