YES

We show the termination of the TRS R:

  -(x,|0|()) -> x
  -(|0|(),s(y)) -> |0|()
  -(s(x),s(y)) -> -(x,y)
  f(|0|()) -> |0|()
  f(s(x)) -> -(s(x),g(f(x)))
  g(|0|()) -> s(|0|())
  g(s(x)) -> -(s(x),f(g(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)) -> -#(s(x),g(f(x)))
p3: f#(s(x)) -> g#(f(x))
p4: f#(s(x)) -> f#(x)
p5: g#(s(x)) -> -#(s(x),f(g(x)))
p6: g#(s(x)) -> f#(g(x))
p7: g#(s(x)) -> g#(x)

and R consists of:

r1: -(x,|0|()) -> x
r2: -(|0|(),s(y)) -> |0|()
r3: -(s(x),s(y)) -> -(x,y)
r4: f(|0|()) -> |0|()
r5: f(s(x)) -> -(s(x),g(f(x)))
r6: g(|0|()) -> s(|0|())
r7: g(s(x)) -> -(s(x),f(g(x)))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p6, p7}
  {p1}


-- Reduction pair.

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

p1: g#(s(x)) -> g#(x)
p2: g#(s(x)) -> f#(g(x))
p3: f#(s(x)) -> f#(x)
p4: f#(s(x)) -> g#(f(x))

and R consists of:

r1: -(x,|0|()) -> x
r2: -(|0|(),s(y)) -> |0|()
r3: -(s(x),s(y)) -> -(x,y)
r4: f(|0|()) -> |0|()
r5: f(s(x)) -> -(s(x),g(f(x)))
r6: g(|0|()) -> s(|0|())
r7: g(s(x)) -> -(s(x),f(g(x)))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      g#_A(x1) = x1 + (2,0)
      s_A(x1) = x1 + (5,2)
      f#_A(x1) = ((1,0),(1,0)) x1
      g_A(x1) = x1 + (6,1)
      f_A(x1) = ((1,0),(1,0)) x1 + (2,1)
      -_A(x1,x2) = x1 + (1,5)
      |0|_A() = (1,3)

The next rules are strictly ordered:

  p1, p2, p3, p4

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: -(|0|(),s(y)) -> |0|()
r3: -(s(x),s(y)) -> -(x,y)
r4: f(|0|()) -> |0|()
r5: f(s(x)) -> -(s(x),g(f(x)))
r6: g(|0|()) -> s(|0|())
r7: g(s(x)) -> -(s(x),f(g(x)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7

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