YES

We show the termination of the TRS R:

  minus(minus(x)) -> x
  minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
  minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
  f(minus(x)) -> minus(minus(minus(f(x))))

-- SCC decomposition.

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

p1: minus#(+(x,y)) -> minus#(minus(minus(x)))
p2: minus#(+(x,y)) -> minus#(minus(x))
p3: minus#(+(x,y)) -> minus#(x)
p4: minus#(+(x,y)) -> minus#(minus(minus(y)))
p5: minus#(+(x,y)) -> minus#(minus(y))
p6: minus#(+(x,y)) -> minus#(y)
p7: minus#(*(x,y)) -> minus#(minus(minus(x)))
p8: minus#(*(x,y)) -> minus#(minus(x))
p9: minus#(*(x,y)) -> minus#(x)
p10: minus#(*(x,y)) -> minus#(minus(minus(y)))
p11: minus#(*(x,y)) -> minus#(minus(y))
p12: minus#(*(x,y)) -> minus#(y)
p13: f#(minus(x)) -> minus#(minus(minus(f(x))))
p14: f#(minus(x)) -> minus#(minus(f(x)))
p15: f#(minus(x)) -> minus#(f(x))
p16: f#(minus(x)) -> f#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The estimated dependency graph contains the following SCCs:

  {p16}
  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

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

p1: f#(minus(x)) -> f#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      f#_A(x1) = ((1,0),(1,0)) x1
      minus_A(x1) = ((1,1),(1,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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#(minus(minus(x)))
p2: minus#(*(x,y)) -> minus#(y)
p3: minus#(*(x,y)) -> minus#(minus(y))
p4: minus#(*(x,y)) -> minus#(minus(minus(y)))
p5: minus#(*(x,y)) -> minus#(x)
p6: minus#(*(x,y)) -> minus#(minus(x))
p7: minus#(*(x,y)) -> minus#(minus(minus(x)))
p8: minus#(+(x,y)) -> minus#(y)
p9: minus#(+(x,y)) -> minus#(minus(y))
p10: minus#(+(x,y)) -> minus#(minus(minus(y)))
p11: minus#(+(x,y)) -> minus#(x)
p12: minus#(+(x,y)) -> minus#(minus(x))

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(+(x,y)) -> *(minus(minus(minus(x))),minus(minus(minus(y))))
r3: minus(*(x,y)) -> +(minus(minus(minus(x))),minus(minus(minus(y))))
r4: f(minus(x)) -> minus(minus(minus(f(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      minus#_A(x1) = ((0,1),(0,0)) x1
      +_A(x1,x2) = x1 + x2 + (1,1)
      minus_A(x1) = ((1,1),(0,1)) x1 + (1,0)
      *_A(x1,x2) = x1 + x2 + (1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12

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