YES
We show the termination of the TRS R:

  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  quot(|0|(),s(y)) -> |0|()
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
  minus(minus(x,y),z) -> minus(x,plus(y,z))
  plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
  plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
  plus(zero(),y) -> y
  plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
  id(x) -> x
  if(true(),x,y) -> x
  if(false(),x,y) -> y
  not(x) -> if(x,false(),true())
  gt(s(x),zero()) -> true()
  gt(zero(),y) -> false()
  gt(s(x),s(y)) -> gt(x,y)


-- SCC decomposition.

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

p1: minus#(s(x),s(y)) -> minus#(x,y)
p2: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
p3: quot#(s(x),s(y)) -> minus#(x,y)
p4: minus#(minus(x,y),z) -> minus#(x,plus(y,z))
p5: minus#(minus(x,y),z) -> plus#(y,z)
p6: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p7: plus#(s(x),s(y)) -> if#(gt(x,y),x,y)
p8: plus#(s(x),s(y)) -> gt#(x,y)
p9: plus#(s(x),s(y)) -> if#(not(gt(x,y)),id(x),id(y))
p10: plus#(s(x),s(y)) -> not#(gt(x,y))
p11: plus#(s(x),s(y)) -> id#(x)
p12: plus#(s(x),s(y)) -> id#(y)
p13: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))
p14: plus#(s(x),x) -> if#(gt(x,x),id(x),id(x))
p15: plus#(s(x),x) -> gt#(x,x)
p16: plus#(s(x),x) -> id#(x)
p17: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p18: plus#(id(x),s(y)) -> if#(gt(s(y),y),y,s(y))
p19: plus#(id(x),s(y)) -> gt#(s(y),y)
p20: not#(x) -> if#(x,false(),true())
p21: gt#(s(x),s(y)) -> gt#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: minus(minus(x,y),z) -> minus(x,plus(y,z))
r6: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r7: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r8: plus(zero(),y) -> y
r9: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r10: id(x) -> x
r11: if(true(),x,y) -> x
r12: if(false(),x,y) -> y
r13: not(x) -> if(x,false(),true())
r14: gt(s(x),zero()) -> true()
r15: gt(zero(),y) -> false()
r16: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1, p4}
  {p6, p13, p17}
  {p21}


-- Reduction pair.

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

p1: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: minus(minus(x,y),z) -> minus(x,plus(y,z))
r6: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r7: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r8: plus(zero(),y) -> y
r9: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r10: id(x) -> x
r11: if(true(),x,y) -> x
r12: if(false(),x,y) -> y
r13: not(x) -> if(x,false(),true())
r14: gt(s(x),zero()) -> true()
r15: gt(zero(),y) -> false()
r16: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r1, r2, r5, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = 0
            s_A(x1) = x1 + 51
            minus_A(x1,x2) = x1 + 12
            |0|_A = 13
            plus_A(x1,x2) = 1
            gt_A(x1,x2) = max{1, x1 - 4}
            zero_A = 3
            true_A = 10
            false_A = 1
            quot#!_A(x1,x2) = x1 + 56
            s!_A(x1) = 68
            minus!_A(x1,x2) = 0
            |0|!_A = 2
            plus!_A(x1,x2) = max{28, x1 + 16, x2 + 19}
            gt!_A(x1,x2) = max{33, x1 - 52}
            zero!_A = 2
            true!_A = 1
            false!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x2 - 28}
            s_A(x1) = max{51, x1 + 28}
            minus_A(x1,x2) = x1 + 50
            |0|_A = 2
            plus_A(x1,x2) = max{1, x2 - 49}
            gt_A(x1,x2) = max{40, x1 - 16, x2 + 34}
            zero_A = 0
            true_A = 36
            false_A = 35
            quot#!_A(x1,x2) = 1
            s!_A(x1) = 0
            minus!_A(x1,x2) = max{27, x1 - 49}
            |0|!_A = 73
            plus!_A(x1,x2) = max{72, x1 + 3, x2 - 25}
            gt!_A(x1,x2) = max{73, x2 + 23}
            zero!_A = 25
            true!_A = 0
            false!_A = 24
        
  
    partial status:
  
      pi(quot#) = []
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(plus) = [1, 2]
      pi(gt) = []
      pi(zero) = []
      pi(true) = []
      pi(false) = []

The following rules are removed:

  p1

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)
p2: minus#(minus(x,y),z) -> minus#(x,plus(y,z))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: minus(minus(x,y),z) -> minus(x,plus(y,z))
r6: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r7: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r8: plus(zero(),y) -> y
r9: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r10: id(x) -> x
r11: if(true(),x,y) -> x
r12: if(false(),x,y) -> y
r13: not(x) -> if(x,false(),true())
r14: gt(s(x),zero()) -> true()
r15: gt(zero(),y) -> false()
r16: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = 0
            s_A(x1) = max{25, x1 + 6}
            minus_A(x1,x2) = x2 + 2
            plus_A(x1,x2) = max{103, x1 + 36, x2 - 6}
            minus#!_A(x1,x2) = max{3, x1 + 1}
            s!_A(x1) = max{59, x1 - 42}
            minus!_A(x1,x2) = x1 + 41
            plus!_A(x1,x2) = 101
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = 0
            s_A(x1) = 85
            minus_A(x1,x2) = 1
            plus_A(x1,x2) = x1 + 42
            minus#!_A(x1,x2) = 1
            s!_A(x1) = x1 + 11
            minus!_A(x1,x2) = max{2, x1 - 72, x2 - 187}
            plus!_A(x1,x2) = max{56, x1 + 51, x2 - 87}
        
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(plus) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

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

p1: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p2: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p3: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: minus(minus(x,y),z) -> minus(x,plus(y,z))
r6: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r7: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r8: plus(zero(),y) -> y
r9: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r10: id(x) -> x
r11: if(true(),x,y) -> x
r12: if(false(),x,y) -> y
r13: not(x) -> if(x,false(),true())
r14: gt(s(x),zero()) -> true()
r15: gt(zero(),y) -> false()
r16: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r10, r11, r12, r13, r14, r15, r16

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = 7
            if_A(x1,x2,x3) = max{14, x1 - 18, x2, x3 + 10}
            gt_A(x1,x2) = 18
            not_A(x1) = max{14, x1}
            id_A(x1) = x1
            true_A = 1
            false_A = 1
            zero_A = 0
            plus#!_A(x1,x2) = 0
            s!_A(x1) = max{22, x1 + 4}
            if!_A(x1,x2,x3) = max{5, x1 - 18, x2 - 3, x3 - 4}
            gt!_A(x1,x2) = max{7, x2}
            not!_A(x1) = max{23, x1 - 16}
            id!_A(x1) = 0
            true!_A = 5
            false!_A = 1
            zero!_A = 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 87, x2 - 87}
            s_A(x1) = x1 + 16
            if_A(x1,x2,x3) = max{2, x1 - 31, x2, x3}
            gt_A(x1,x2) = x2 + 44
            not_A(x1) = max{50, x1 + 3}
            id_A(x1) = max{1, x1}
            true_A = 27
            false_A = 4
            zero_A = 0
            plus#!_A(x1,x2) = max{x1 + 6, x2 + 3}
            s!_A(x1) = max{50, x1 - 29}
            if!_A(x1,x2,x3) = max{51, x1 - 1, x2 - 63, x3 - 28}
            gt!_A(x1,x2) = max{50, x1 - 45, x2 - 56}
            not!_A(x1) = max{1, x1 - 102}
            id!_A(x1) = max{23, x1 - 1}
            true!_A = 50
            false!_A = 52
            zero!_A = 51
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = [1, 2, 3]
      pi(gt) = [2]
      pi(not) = [1]
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p3



-- SCC decomposition.

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

p1: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p2: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: minus(minus(x,y),z) -> minus(x,plus(y,z))
r6: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r7: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r8: plus(zero(),y) -> y
r9: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r10: id(x) -> x
r11: if(true(),x,y) -> x
r12: if(false(),x,y) -> y
r13: not(x) -> if(x,false(),true())
r14: gt(s(x),zero()) -> true()
r15: gt(zero(),y) -> false()
r16: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p2: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: minus(minus(x,y),z) -> minus(x,plus(y,z))
r6: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r7: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r8: plus(zero(),y) -> y
r9: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r10: id(x) -> x
r11: if(true(),x,y) -> x
r12: if(false(),x,y) -> y
r13: not(x) -> if(x,false(),true())
r14: gt(s(x),zero()) -> true()
r15: gt(zero(),y) -> false()
r16: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r10, r11, r12, r13, r14, r15, r16

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 113, x2 - 111}
            s_A(x1) = max{44, x1 + 31}
            if_A(x1,x2,x3) = max{77, x1 - 78, x2 + 29, x3}
            gt_A(x1,x2) = max{188, x1 - 45, x2 - 119}
            not_A(x1) = max{139, x1 - 78}
            id_A(x1) = max{45, x1}
            true_A = 1
            false_A = 0
            zero_A = 0
            plus#!_A(x1,x2) = max{353, x1 + 231, x2 + 200}
            s!_A(x1) = max{191, x1 + 33}
            if!_A(x1,x2,x3) = max{352, x1 + 2, x2 - 95, x3 - 45}
            gt!_A(x1,x2) = max{230, x1 - 45, x2 - 85}
            not!_A(x1) = max{354, x1 + 351}
            id!_A(x1) = 0
            true!_A = 190
            false!_A = 229
            zero!_A = 230
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 62, x2 - 42}
            s_A(x1) = x1 + 22
            if_A(x1,x2,x3) = max{x1 - 34, x2, x3}
            gt_A(x1,x2) = max{11, x1 - 45, x2 - 23}
            not_A(x1) = max{13, x1 - 34}
            id_A(x1) = max{41, x1}
            true_A = 11
            false_A = 1
            zero_A = 0
            plus#!_A(x1,x2) = max{79, x1 + 58, x2 + 37}
            s!_A(x1) = 101
            if!_A(x1,x2,x3) = max{98, x1 - 14, x2 - 2, x3 - 23}
            gt!_A(x1,x2) = max{100, x1 - 1, x2 - 71}
            not!_A(x1) = max{100, x1 - 13}
            id!_A(x1) = max{1, x1 - 71}
            true!_A = 22
            false!_A = 0
            zero!_A = 23
        
  
    partial status:
  
      pi(plus#) = [1, 2]
      pi(s) = [1]
      pi(if) = [2, 3]
      pi(gt) = [2]
      pi(not) = [1]
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

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

p1: gt#(s(x),s(y)) -> gt#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: minus(minus(x,y),z) -> minus(x,plus(y,z))
r6: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r7: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r8: plus(zero(),y) -> y
r9: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r10: id(x) -> x
r11: if(true(),x,y) -> x
r12: if(false(),x,y) -> y
r13: not(x) -> if(x,false(),true())
r14: gt(s(x),zero()) -> true()
r15: gt(zero(),y) -> false()
r16: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = max{0, x1 - 1, x2 - 1}
            s_A(x1) = max{18, x1}
            gt#!_A(x1,x2) = max{1, x1 - 19, x2 - 19}
            s!_A(x1) = max{3, x1 - 50}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = max{0, x1 - 1, x2 - 1}
            s_A(x1) = max{5, x1}
            gt#!_A(x1,x2) = max{1, x1 - 6, x2 - 6}
            s!_A(x1) = 0
        
  
    partial status:
  
      pi(gt#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.