YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: +#(x,minus(y)) -> minus#(+(minus(x),y))
p2: +#(x,minus(y)) -> +#(minus(x),y)
p3: +#(x,minus(y)) -> minus#(x)
p4: +#(x,+(y,z)) -> +#(+(x,y),z)
p5: +#(x,+(y,z)) -> +#(x,y)
p6: +#(minus(+(x,|1|())),|1|()) -> minus#(x)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p4, p5}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        +#_A(x1,x2) = max{x1 + 3, x2 + 7}
        minus_A(x1) = x1
        +_A(x1,x2) = max{x1, x2 + 3}
        |0|_A = 3
        |1|_A = 4
  
    precedence:
    
      minus = + > |1| > +# = |0|
    
    partial status:
  
      pi(+#) = [2]
      pi(minus) = []
      pi(+) = []
      pi(|0|) = []
      pi(|1|) = []

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

and R consists of:

r1: minus(|0|()) -> |0|()
r2: +(x,|0|()) -> x
r3: +(|0|(),y) -> y
r4: +(minus(|1|()),|1|()) -> |0|()
r5: minus(minus(x)) -> x
r6: +(x,minus(y)) -> minus(+(minus(x),y))
r7: +(x,+(y,z)) -> +(+(x,y),z)
r8: +(minus(+(x,|1|())),|1|()) -> minus(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: +#(x,minus(y)) -> +#(minus(x),y)
p2: +#(x,+(y,z)) -> +#(x,y)

and R consists of:

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

The set of usable rules consists of

  r1, r5

Take the reduction pair:

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

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,y)

and R consists of:

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

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,y)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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