Term Rewriting System R:
[x, y, u, z]
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
<='(s(x), s(y)) -> <='(x, y)
PERFECTP(s(x)) -> F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)
F(s(x), 0, z, u) -> -'(z, s(x))
F(s(x), s(y), z, u) -> IF(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))
F(s(x), s(y), z, u) -> <='(x, y)
F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)
F(s(x), s(y), z, u) -> -'(y, x)
F(s(x), s(y), z, u) -> F(x, u, z, u)

Furthermore, R contains three SCCs.

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

Dependency Pair:

-'(s(x), s(y)) -> -'(x, y)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The following dependency pair can be strictly oriented:

-'(s(x), s(y)) -> -'(x, y)

Additionally, the following rules can be oriented:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(if(x1, x2, x3)) =  x2 + x3 POL(0) =  0 POL(-'(x1, x2)) =  x1 POL(false) =  0 POL(perfectp(x1)) =  0 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(-(x1, x2)) =  x1 POL(f(x1, x2, x3, x4)) =  0 POL(<=(x1, x2)) =  0

resulting in one new DP problem.

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

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

<='(s(x), s(y)) -> <='(x, y)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The following dependency pair can be strictly oriented:

<='(s(x), s(y)) -> <='(x, y)

Additionally, the following rules can be oriented:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(if(x1, x2, x3)) =  x2 + x3 POL(0) =  0 POL(false) =  0 POL(perfectp(x1)) =  0 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(-(x1, x2)) =  x1 POL(<='(x1, x2)) =  x1 POL(f(x1, x2, x3, x4)) =  0 POL(<=(x1, x2)) =  0

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 5
Dependency Graph
→DP Problem 3
Polo

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pairs:

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)
F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The following dependency pairs can be strictly oriented:

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)

Additionally, the following rules can be oriented:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(if(x1, x2, x3)) =  x2 + x3 POL(0) =  0 POL(false) =  0 POL(perfectp(x1)) =  0 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(-(x1, x2)) =  x1 POL(f(x1, x2, x3, x4)) =  0 POL(F(x1, x2, x3, x4)) =  x1 POL(<=(x1, x2)) =  0

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 6
Polynomial Ordering

Dependency Pair:

F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The following dependency pair can be strictly oriented:

F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)

Additionally, the following rules can be oriented:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(if(x1, x2, x3)) =  x2 + x3 POL(0) =  0 POL(false) =  0 POL(perfectp(x1)) =  0 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(-(x1, x2)) =  x1 POL(f(x1, x2, x3, x4)) =  0 POL(F(x1, x2, x3, x4)) =  x2 POL(<=(x1, x2)) =  0

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 6
Polo
...
→DP Problem 7
Dependency Graph

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Using the Dependency Graph resulted in no new DP problems.

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