YES

We show the termination of the TRS R:

  -(x,|0|()) -> x
  -(s(x),s(y)) -> -(x,y)
  p(s(x)) -> x
  f(s(x),y) -> f(p(-(s(x),y)),p(-(y,s(x))))
  f(x,s(y)) -> f(p(-(x,s(y))),p(-(s(y),x)))

-- SCC decomposition.

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

p1: -#(s(x),s(y)) -> -#(x,y)
p2: f#(s(x),y) -> f#(p(-(s(x),y)),p(-(y,s(x))))
p3: f#(s(x),y) -> p#(-(s(x),y))
p4: f#(s(x),y) -> -#(s(x),y)
p5: f#(s(x),y) -> p#(-(y,s(x)))
p6: f#(s(x),y) -> -#(y,s(x))
p7: f#(x,s(y)) -> f#(p(-(x,s(y))),p(-(s(y),x)))
p8: f#(x,s(y)) -> p#(-(x,s(y)))
p9: f#(x,s(y)) -> -#(x,s(y))
p10: f#(x,s(y)) -> p#(-(s(y),x))
p11: f#(x,s(y)) -> -#(s(y),x)

and R consists of:

r1: -(x,|0|()) -> x
r2: -(s(x),s(y)) -> -(x,y)
r3: p(s(x)) -> x
r4: f(s(x),y) -> f(p(-(s(x),y)),p(-(y,s(x))))
r5: f(x,s(y)) -> f(p(-(x,s(y))),p(-(s(y),x)))

The estimated dependency graph contains the following SCCs:

  {p2, p7}
  {p1}


-- Reduction pair.

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

p1: f#(x,s(y)) -> f#(p(-(x,s(y))),p(-(s(y),x)))
p2: f#(s(x),y) -> f#(p(-(s(x),y)),p(-(y,s(x))))

and R consists of:

r1: -(x,|0|()) -> x
r2: -(s(x),s(y)) -> -(x,y)
r3: p(s(x)) -> x
r4: f(s(x),y) -> f(p(-(s(x),y)),p(-(y,s(x))))
r5: f(x,s(y)) -> f(p(-(x,s(y))),p(-(s(y),x)))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        p > f# > s > |0| > -
      
      argument filter:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(p) = 1
        pi(-) = 1
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2) = x1
        s_A(x1) = 1
        p_A(x1) = 0
        -_A(x1,x2) = x1 + 1
        |0|_A() = 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: f#(x,s(y)) -> f#(p(-(x,s(y))),p(-(s(y),x)))

and R consists of:

r1: -(x,|0|()) -> x
r2: -(s(x),s(y)) -> -(x,y)
r3: p(s(x)) -> x
r4: f(s(x),y) -> f(p(-(s(x),y)),p(-(y,s(x))))
r5: f(x,s(y)) -> f(p(-(x,s(y))),p(-(s(y),x)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(x,s(y)) -> f#(p(-(x,s(y))),p(-(s(y),x)))

and R consists of:

r1: -(x,|0|()) -> x
r2: -(s(x),s(y)) -> -(x,y)
r3: p(s(x)) -> x
r4: f(s(x),y) -> f(p(-(s(x),y)),p(-(y,s(x))))
r5: f(x,s(y)) -> f(p(-(x,s(y))),p(-(s(y),x)))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > - > f# > s > p
      
      argument filter:
    
        pi(f#) = [2]
        pi(s) = [1]
        pi(p) = 1
        pi(-) = 1
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2) = x2
        s_A(x1) = x1 + 2
        p_A(x1) = 1
        -_A(x1,x2) = x1 + 1
        |0|_A() = 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: -#(s(x),s(y)) -> -#(x,y)

and R consists of:

r1: -(x,|0|()) -> x
r2: -(s(x),s(y)) -> -(x,y)
r3: p(s(x)) -> x
r4: f(s(x),y) -> f(p(-(s(x),y)),p(-(y,s(x))))
r5: f(x,s(y)) -> f(p(-(x,s(y))),p(-(s(y),x)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > -#
      
      argument filter:
    
        pi(-#) = 2
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        -#_A(x1,x2) = 0
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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