YES

We show the termination of the TRS R:

  minus(minus(x)) -> x
  minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
  minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
  f(minus(x)) -> minus(minus(minus(f(x))))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(minus(x)))
p2: minus#(+(x,y)) -> minus#(minus(x))
p3: minus#(+(x,y)) -> minus#(x)
p4: minus#(+(x,y)) -> minus#(minus(minus(y)))
p5: minus#(+(x,y)) -> minus#(minus(y))
p6: minus#(+(x,y)) -> minus#(y)
p7: minus#(*(x,y)) -> minus#(minus(minus(x)))
p8: minus#(*(x,y)) -> minus#(minus(x))
p9: minus#(*(x,y)) -> minus#(x)
p10: minus#(*(x,y)) -> minus#(minus(minus(y)))
p11: minus#(*(x,y)) -> minus#(minus(y))
p12: minus#(*(x,y)) -> minus#(y)
p13: f#(minus(x)) -> minus#(minus(minus(f(x))))
p14: f#(minus(x)) -> minus#(minus(f(x)))
p15: f#(minus(x)) -> minus#(f(x))
p16: f#(minus(x)) -> f#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p16}
  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(minus(x)) -> f#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = x1 + 2
        minus_A(x1) = x1 + 2
  
    precedence:
    
      f# = minus
    
    partial status:
  
      pi(f#) = [1]
      pi(minus) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(minus(x)))
p2: minus#(*(x,y)) -> minus#(y)
p3: minus#(*(x,y)) -> minus#(minus(y))
p4: minus#(*(x,y)) -> minus#(minus(minus(y)))
p5: minus#(*(x,y)) -> minus#(x)
p6: minus#(*(x,y)) -> minus#(minus(x))
p7: minus#(*(x,y)) -> minus#(minus(minus(x)))
p8: minus#(+(x,y)) -> minus#(y)
p9: minus#(+(x,y)) -> minus#(minus(y))
p10: minus#(+(x,y)) -> minus#(minus(minus(y)))
p11: minus#(+(x,y)) -> minus#(x)
p12: minus#(+(x,y)) -> minus#(minus(x))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = x1 + 3
        +_A(x1,x2) = max{x1 + 3, x2 + 4}
        minus_A(x1) = x1
        *_A(x1,x2) = max{x1 + 3, x2 + 4}
  
    precedence:
    
      + = * > minus# = minus
    
    partial status:
  
      pi(minus#) = [1]
      pi(+) = []
      pi(minus) = [1]
      pi(*) = []

The next rules are strictly ordered:

  p10

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(minus(x)))
p2: minus#(*(x,y)) -> minus#(y)
p3: minus#(*(x,y)) -> minus#(minus(y))
p4: minus#(*(x,y)) -> minus#(minus(minus(y)))
p5: minus#(*(x,y)) -> minus#(x)
p6: minus#(*(x,y)) -> minus#(minus(x))
p7: minus#(*(x,y)) -> minus#(minus(minus(x)))
p8: minus#(+(x,y)) -> minus#(y)
p9: minus#(+(x,y)) -> minus#(minus(y))
p10: minus#(+(x,y)) -> minus#(x)
p11: minus#(+(x,y)) -> minus#(minus(x))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(minus(x)))
p2: minus#(+(x,y)) -> minus#(minus(x))
p3: minus#(+(x,y)) -> minus#(x)
p4: minus#(+(x,y)) -> minus#(minus(y))
p5: minus#(+(x,y)) -> minus#(y)
p6: minus#(*(x,y)) -> minus#(minus(minus(x)))
p7: minus#(*(x,y)) -> minus#(minus(x))
p8: minus#(*(x,y)) -> minus#(x)
p9: minus#(*(x,y)) -> minus#(minus(minus(y)))
p10: minus#(*(x,y)) -> minus#(minus(y))
p11: minus#(*(x,y)) -> minus#(y)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = max{0, x1 - 2}
        +_A(x1,x2) = max{3, x1 + 2, x2}
        minus_A(x1) = max{1, x1}
        *_A(x1,x2) = max{3, x1 + 2, x2}
  
    precedence:
    
      + = minus > * > minus#
    
    partial status:
  
      pi(minus#) = []
      pi(+) = [1]
      pi(minus) = [1]
      pi(*) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(x))
p2: minus#(+(x,y)) -> minus#(x)
p3: minus#(+(x,y)) -> minus#(minus(y))
p4: minus#(+(x,y)) -> minus#(y)
p5: minus#(*(x,y)) -> minus#(minus(minus(x)))
p6: minus#(*(x,y)) -> minus#(minus(x))
p7: minus#(*(x,y)) -> minus#(x)
p8: minus#(*(x,y)) -> minus#(minus(minus(y)))
p9: minus#(*(x,y)) -> minus#(minus(y))
p10: minus#(*(x,y)) -> minus#(y)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(x))
p2: minus#(*(x,y)) -> minus#(y)
p3: minus#(*(x,y)) -> minus#(minus(y))
p4: minus#(*(x,y)) -> minus#(minus(minus(y)))
p5: minus#(*(x,y)) -> minus#(x)
p6: minus#(*(x,y)) -> minus#(minus(x))
p7: minus#(*(x,y)) -> minus#(minus(minus(x)))
p8: minus#(+(x,y)) -> minus#(y)
p9: minus#(+(x,y)) -> minus#(minus(y))
p10: minus#(+(x,y)) -> minus#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = x1 + 4
        +_A(x1,x2) = max{x1 + 3, x2 + 3}
        minus_A(x1) = x1
        *_A(x1,x2) = max{x1 + 3, x2 + 3}
  
    precedence:
    
      minus# = + = minus = *
    
    partial status:
  
      pi(minus#) = []
      pi(+) = []
      pi(minus) = [1]
      pi(*) = [2]

The next rules are strictly ordered:

  p10

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(x))
p2: minus#(*(x,y)) -> minus#(y)
p3: minus#(*(x,y)) -> minus#(minus(y))
p4: minus#(*(x,y)) -> minus#(minus(minus(y)))
p5: minus#(*(x,y)) -> minus#(x)
p6: minus#(*(x,y)) -> minus#(minus(x))
p7: minus#(*(x,y)) -> minus#(minus(minus(x)))
p8: minus#(+(x,y)) -> minus#(y)
p9: minus#(+(x,y)) -> minus#(minus(y))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(x))
p2: minus#(+(x,y)) -> minus#(minus(y))
p3: minus#(+(x,y)) -> minus#(y)
p4: minus#(*(x,y)) -> minus#(minus(minus(x)))
p5: minus#(*(x,y)) -> minus#(minus(x))
p6: minus#(*(x,y)) -> minus#(x)
p7: minus#(*(x,y)) -> minus#(minus(minus(y)))
p8: minus#(*(x,y)) -> minus#(minus(y))
p9: minus#(*(x,y)) -> minus#(y)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = x1 + 4
        +_A(x1,x2) = max{9, x1 + 3, x2 + 7}
        minus_A(x1) = max{2, x1}
        *_A(x1,x2) = max{9, x1 + 3, x2 + 7}
  
    precedence:
    
      + = minus > minus# = *
    
    partial status:
  
      pi(minus#) = [1]
      pi(+) = [1, 2]
      pi(minus) = [1]
      pi(*) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(+(x,y)) -> minus#(y)
p3: minus#(*(x,y)) -> minus#(minus(minus(x)))
p4: minus#(*(x,y)) -> minus#(minus(x))
p5: minus#(*(x,y)) -> minus#(x)
p6: minus#(*(x,y)) -> minus#(minus(minus(y)))
p7: minus#(*(x,y)) -> minus#(minus(y))
p8: minus#(*(x,y)) -> minus#(y)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(y)
p3: minus#(*(x,y)) -> minus#(minus(y))
p4: minus#(*(x,y)) -> minus#(minus(minus(y)))
p5: minus#(*(x,y)) -> minus#(x)
p6: minus#(*(x,y)) -> minus#(minus(x))
p7: minus#(*(x,y)) -> minus#(minus(minus(x)))
p8: minus#(+(x,y)) -> minus#(y)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = max{4, x1}
        +_A(x1,x2) = max{x1 + 2, x2 + 4}
        minus_A(x1) = x1
        *_A(x1,x2) = max{x1 + 2, x2 + 4}
  
    precedence:
    
      + = * > minus# = minus
    
    partial status:
  
      pi(minus#) = [1]
      pi(+) = []
      pi(minus) = [1]
      pi(*) = []

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(y)
p3: minus#(*(x,y)) -> minus#(minus(y))
p4: minus#(*(x,y)) -> minus#(minus(minus(y)))
p5: minus#(*(x,y)) -> minus#(x)
p6: minus#(*(x,y)) -> minus#(minus(x))
p7: minus#(*(x,y)) -> minus#(minus(minus(x)))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(minus(minus(x)))
p3: minus#(*(x,y)) -> minus#(minus(x))
p4: minus#(*(x,y)) -> minus#(x)
p5: minus#(*(x,y)) -> minus#(minus(minus(y)))
p6: minus#(*(x,y)) -> minus#(minus(y))
p7: minus#(*(x,y)) -> minus#(y)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = max{4, x1 + 3}
        +_A(x1,x2) = max{x1 + 6, x2 + 6}
        minus_A(x1) = x1
        *_A(x1,x2) = max{x1 + 6, x2 + 6}
  
    precedence:
    
      + = minus > minus# > *
    
    partial status:
  
      pi(minus#) = [1]
      pi(+) = []
      pi(minus) = [1]
      pi(*) = [2]

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(minus(minus(x)))
p3: minus#(*(x,y)) -> minus#(minus(x))
p4: minus#(*(x,y)) -> minus#(x)
p5: minus#(*(x,y)) -> minus#(minus(minus(y)))
p6: minus#(*(x,y)) -> minus#(minus(y))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(minus(y))
p3: minus#(*(x,y)) -> minus#(minus(minus(y)))
p4: minus#(*(x,y)) -> minus#(x)
p5: minus#(*(x,y)) -> minus#(minus(x))
p6: minus#(*(x,y)) -> minus#(minus(minus(x)))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = x1 + 3
        +_A(x1,x2) = max{x1 + 4, x2 + 4}
        minus_A(x1) = x1
        *_A(x1,x2) = max{x1 + 4, x2 + 4}
  
    precedence:
    
      minus# = + = minus = *
    
    partial status:
  
      pi(minus#) = [1]
      pi(+) = [1, 2]
      pi(minus) = [1]
      pi(*) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(minus(minus(y)))
p3: minus#(*(x,y)) -> minus#(x)
p4: minus#(*(x,y)) -> minus#(minus(x))
p5: minus#(*(x,y)) -> minus#(minus(minus(x)))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(minus(minus(x)))
p3: minus#(*(x,y)) -> minus#(minus(x))
p4: minus#(*(x,y)) -> minus#(x)
p5: minus#(*(x,y)) -> minus#(minus(minus(y)))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = max{3, x1 + 2}
        +_A(x1,x2) = max{x1 + 5, x2 + 2}
        minus_A(x1) = x1
        *_A(x1,x2) = max{x1 + 5, x2 + 2}
  
    precedence:
    
      + = minus > * > minus#
    
    partial status:
  
      pi(minus#) = [1]
      pi(+) = []
      pi(minus) = []
      pi(*) = [1]

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(minus(minus(x)))
p3: minus#(*(x,y)) -> minus#(minus(x))
p4: minus#(*(x,y)) -> minus#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(x)
p3: minus#(*(x,y)) -> minus#(minus(x))
p4: minus#(*(x,y)) -> minus#(minus(minus(x)))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = max{2, x1}
        +_A(x1,x2) = max{x1 + 2, x2 + 2}
        minus_A(x1) = x1
        *_A(x1,x2) = max{x1 + 2, x2 + 2}
  
    precedence:
    
      + = * > minus# = minus
    
    partial status:
  
      pi(minus#) = [1]
      pi(+) = []
      pi(minus) = [1]
      pi(*) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(x)
p3: minus#(*(x,y)) -> minus#(minus(x))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(minus(x))
p3: minus#(*(x,y)) -> minus#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = x1 + 4
        +_A(x1,x2) = max{x1 + 4, x2 + 4}
        minus_A(x1) = x1
        *_A(x1,x2) = max{x1 + 4, x2 + 4}
  
    precedence:
    
      + = * > minus > minus#
    
    partial status:
  
      pi(minus#) = []
      pi(+) = []
      pi(minus) = [1]
      pi(*) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(minus(x))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))
p2: minus#(*(x,y)) -> minus#(minus(x))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = max{8, x1 + 6}
        +_A(x1,x2) = max{6, x1 + 3, x2 + 5}
        minus_A(x1) = max{1, x1}
        *_A(x1,x2) = max{6, x1 + 3, x2 + 5}
  
    precedence:
    
      minus# = + = minus = *
    
    partial status:
  
      pi(minus#) = []
      pi(+) = []
      pi(minus) = [1]
      pi(*) = [2]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(+(x,y)) -> minus#(minus(y))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1) = max{7, x1 + 6}
        +_A(x1,x2) = max{6, x1 + 1, x2 + 4}
        minus_A(x1) = max{2, x1}
        *_A(x1,x2) = max{6, x1 + 1, x2 + 4}
  
    precedence:
    
      minus# > + = minus = *
    
    partial status:
  
      pi(minus#) = []
      pi(+) = [1, 2]
      pi(minus) = [1]
      pi(*) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.