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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = ((0,0,0),(1,0,0),(0,1,0)) x1
      s_A(x1) = x1 + (1,1,1)
      p_A(x1) = ((1,0,0),(0,0,0),(1,0,0)) x1 + (0,0,1)
      -_A(x1,x2) = ((1,0,0),(0,1,0),(1,0,0)) x1 + ((0,0,0),(1,0,0),(0,0,0)) x2 + (0,0,1)
      |0|_A() = (1,1,1)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = ((0,0,0),(1,0,0),(0,1,0)) x2
      s_A(x1) = x1 + (1,2,1)
      p_A(x1) = x1 + (0,1,1)
      -_A(x1,x2) = x1 + (0,1,0)
      |0|_A() = (1,1,1)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      -#_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1
      s_A(x1) = ((1,0,0),(0,0,0),(1,1,0)) x1 + (1,1,1)

The next rules are strictly ordered:

  p1

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