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:

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

The next rules are strictly ordered:

  p2, 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)

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,minus(y)) -> +#(minus(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:

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

The next rules are strictly ordered:

  p1

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