YES

We show the termination of the TRS R:

  +(x,|0|()) -> x
  +(minus(x),x) -> |0|()
  minus(|0|()) -> |0|()
  minus(minus(x)) -> x
  minus(+(x,y)) -> +(minus(y),minus(x))
  *(x,|1|()) -> x
  *(x,|0|()) -> |0|()
  *(x,+(y,z)) -> +(*(x,y),*(x,z))
  *(x,minus(y)) -> minus(*(x,y))

-- SCC decomposition.

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

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,y))

The estimated dependency graph contains the following SCCs:

  {p5, p6, p8}
  {p2, p3}


-- Reduction pair.

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

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          *#_A(x1,x2) = ((0,1),(0,1)) x2 + (2,2)
          minus_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          +_A(x1,x2) = x1 + ((0,0),(0,1)) x2 + (3,2)
  
    precedence:
    
      minus > *# = +
    
    partial status:
  
      pi(*#) = []
      pi(minus) = []
      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: *#(x,+(y,z)) -> *#(x,z)
p2: *#(x,+(y,z)) -> *#(x,y)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,y))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          *#_A(x1,x2) = ((1,0),(0,0)) x2
          +_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (1,1)
  
    precedence:
    
      + > *#
    
    partial status:
  
      pi(*#) = []
      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: *#(x,+(y,z)) -> *#(x,z)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,y))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          *#_A(x1,x2) = x2 + (0,1)
          +_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
  
    precedence:
    
      + > *#
    
    partial status:
  
      pi(*#) = [2]
      pi(+) = [2]

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          minus#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          +_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
  
    precedence:
    
      minus# = +
    
    partial status:
  
      pi(minus#) = []
      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#(x)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,y))

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#(x)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,y))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          minus#_A(x1) = x1 + (1,1)
          +_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (2,1)
  
    precedence:
    
      + > minus#
    
    partial status:
  
      pi(minus#) = [1]
      pi(+) = [1, 2]

The next rules are strictly ordered:

  p1

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