YES

We show the termination of the TRS R:

  norm(nil()) -> |0|()
  norm(g(x,y)) -> s(norm(x))
  f(x,nil()) -> g(nil(),x)
  f(x,g(y,z)) -> g(f(x,y),z)
  rem(nil(),y) -> nil()
  rem(g(x,y),|0|()) -> g(x,y)
  rem(g(x,y),s(z)) -> rem(x,z)

-- SCC decomposition.

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

p1: norm#(g(x,y)) -> norm#(x)
p2: f#(x,g(y,z)) -> f#(x,y)
p3: rem#(g(x,y),s(z)) -> rem#(x,z)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2}
  {p3}


-- Reduction pair.

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

p1: norm#(g(x,y)) -> norm#(x)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      norm#_A(x1) = x1
      g_A(x1,x2) = x1 + 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: f#(x,g(y,z)) -> f#(x,y)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      f#_A(x1,x2) = x2
      g_A(x1,x2) = x1 + 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: rem#(g(x,y),s(z)) -> rem#(x,z)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      rem#_A(x1,x2) = x1 + x2
      g_A(x1,x2) = x1
      s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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