YES

We show the termination of the TRS R:

  minus(|0|(),y) -> |0|()
  minus(s(x),|0|()) -> s(x)
  minus(s(x),s(y)) -> minus(x,y)
  le(|0|(),y) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  if(true(),x,y) -> x
  if(false(),x,y) -> y
  perfectp(|0|()) -> false()
  perfectp(s(x)) -> f(x,s(|0|()),s(x),s(x))
  f(|0|(),y,|0|(),u) -> true()
  f(|0|(),y,s(z),u) -> false()
  f(s(x),|0|(),z,u) -> f(x,u,minus(z,s(x)),u)
  f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))

-- SCC decomposition.

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

p1: minus#(s(x),s(y)) -> minus#(x,y)
p2: le#(s(x),s(y)) -> le#(x,y)
p3: perfectp#(s(x)) -> f#(x,s(|0|()),s(x),s(x))
p4: f#(s(x),|0|(),z,u) -> f#(x,u,minus(z,s(x)),u)
p5: f#(s(x),|0|(),z,u) -> minus#(z,s(x))
p6: f#(s(x),s(y),z,u) -> if#(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))
p7: f#(s(x),s(y),z,u) -> le#(x,y)
p8: f#(s(x),s(y),z,u) -> f#(s(x),minus(y,x),z,u)
p9: f#(s(x),s(y),z,u) -> minus#(y,x)
p10: f#(s(x),s(y),z,u) -> f#(x,u,z,u)

and R consists of:

r1: minus(|0|(),y) -> |0|()
r2: minus(s(x),|0|()) -> s(x)
r3: minus(s(x),s(y)) -> minus(x,y)
r4: le(|0|(),y) -> true()
r5: le(s(x),|0|()) -> false()
r6: le(s(x),s(y)) -> le(x,y)
r7: if(true(),x,y) -> x
r8: if(false(),x,y) -> y
r9: perfectp(|0|()) -> false()
r10: perfectp(s(x)) -> f(x,s(|0|()),s(x),s(x))
r11: f(|0|(),y,|0|(),u) -> true()
r12: f(|0|(),y,s(z),u) -> false()
r13: f(s(x),|0|(),z,u) -> f(x,u,minus(z,s(x)),u)
r14: f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))

The estimated dependency graph contains the following SCCs:

  {p4, p8, p10}
  {p1}
  {p2}


-- Reduction pair.

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

p1: f#(s(x),|0|(),z,u) -> f#(x,u,minus(z,s(x)),u)
p2: f#(s(x),s(y),z,u) -> f#(x,u,z,u)
p3: f#(s(x),s(y),z,u) -> f#(s(x),minus(y,x),z,u)

and R consists of:

r1: minus(|0|(),y) -> |0|()
r2: minus(s(x),|0|()) -> s(x)
r3: minus(s(x),s(y)) -> minus(x,y)
r4: le(|0|(),y) -> true()
r5: le(s(x),|0|()) -> false()
r6: le(s(x),s(y)) -> le(x,y)
r7: if(true(),x,y) -> x
r8: if(false(),x,y) -> y
r9: perfectp(|0|()) -> false()
r10: perfectp(s(x)) -> f(x,s(|0|()),s(x),s(x))
r11: f(|0|(),y,|0|(),u) -> true()
r12: f(|0|(),y,s(z),u) -> false()
r13: f(s(x),|0|(),z,u) -> f(x,u,minus(z,s(x)),u)
r14: f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1,x2,x3,x4) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,1,1,1)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(1,1,1,0)) x2 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,1,1,0)) x4
      s_A(x1) = x1 + (2,2,3,1)
      |0|_A() = (1,2,1,2)
      minus_A(x1,x2) = x1 + ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + (1,1,2,1)

The next rules are strictly ordered:

  p1, p2, p3

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#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: minus(|0|(),y) -> |0|()
r2: minus(s(x),|0|()) -> s(x)
r3: minus(s(x),s(y)) -> minus(x,y)
r4: le(|0|(),y) -> true()
r5: le(s(x),|0|()) -> false()
r6: le(s(x),s(y)) -> le(x,y)
r7: if(true(),x,y) -> x
r8: if(false(),x,y) -> y
r9: perfectp(|0|()) -> false()
r10: perfectp(s(x)) -> f(x,s(|0|()),s(x),s(x))
r11: f(|0|(),y,|0|(),u) -> true()
r12: f(|0|(),y,s(z),u) -> false()
r13: f(s(x),|0|(),z,u) -> f(x,u,minus(z,s(x)),u)
r14: f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      minus#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,1,0,1)) x1 + ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,0,1,1)) x2
      s_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

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: le#(s(x),s(y)) -> le#(x,y)

and R consists of:

r1: minus(|0|(),y) -> |0|()
r2: minus(s(x),|0|()) -> s(x)
r3: minus(s(x),s(y)) -> minus(x,y)
r4: le(|0|(),y) -> true()
r5: le(s(x),|0|()) -> false()
r6: le(s(x),s(y)) -> le(x,y)
r7: if(true(),x,y) -> x
r8: if(false(),x,y) -> y
r9: perfectp(|0|()) -> false()
r10: perfectp(s(x)) -> f(x,s(|0|()),s(x),s(x))
r11: f(|0|(),y,|0|(),u) -> true()
r12: f(|0|(),y,s(z),u) -> false()
r13: f(s(x),|0|(),z,u) -> f(x,u,minus(z,s(x)),u)
r14: f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      le#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,0,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,0,1,1)) x2
      s_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,1,1,1)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

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