YES

We show the termination of the TRS R:

  minus(minus(x)) -> x
  minus(h(x)) -> h(minus(x))
  minus(f(x,y)) -> f(minus(y),minus(x))

-- SCC decomposition.

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

p1: minus#(h(x)) -> minus#(x)
p2: minus#(f(x,y)) -> minus#(y)
p3: minus#(f(x,y)) -> minus#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(h(x)) -> h(minus(x))
r3: minus(f(x,y)) -> f(minus(y),minus(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#(h(x)) -> minus#(x)
p2: minus#(f(x,y)) -> minus#(x)
p3: minus#(f(x,y)) -> minus#(y)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(h(x)) -> h(minus(x))
r3: minus(f(x,y)) -> f(minus(y),minus(x))

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),(0,1)) x1 + (2,2)
          h_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          f_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(0,1)) x2 + (3,1)
  
    precedence:
    
      minus# = h = f
    
    partial status:
  
      pi(minus#) = []
      pi(h) = []
      pi(f) = []

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#(h(x)) -> minus#(x)
p2: minus#(f(x,y)) -> minus#(y)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(h(x)) -> h(minus(x))
r3: minus(f(x,y)) -> f(minus(y),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: minus#(h(x)) -> minus#(x)
p2: minus#(f(x,y)) -> minus#(y)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(h(x)) -> h(minus(x))
r3: minus(f(x,y)) -> f(minus(y),minus(x))

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) = ((1,0),(1,1)) x1 + (2,2)
          h_A(x1) = ((1,1),(0,0)) x1 + (1,1)
          f_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (3,2)
  
    precedence:
    
      f > minus# = h
    
    partial status:
  
      pi(minus#) = []
      pi(h) = []
      pi(f) = [1]

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#(h(x)) -> minus#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(h(x)) -> h(minus(x))
r3: minus(f(x,y)) -> f(minus(y),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: minus#(h(x)) -> minus#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(h(x)) -> h(minus(x))
r3: minus(f(x,y)) -> f(minus(y),minus(x))

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) = ((1,0),(1,0)) x1 + (2,2)
          h_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      minus# = h
    
    partial status:
  
      pi(minus#) = []
      pi(h) = []

The next rules are strictly ordered:

  p1

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