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)))
  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: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p5: plus#(s(x),s(y)) -> if#(gt(x,y),x,y)
p6: plus#(s(x),s(y)) -> gt#(x,y)
p7: plus#(s(x),s(y)) -> if#(not(gt(x,y)),id(x),id(y))
p8: plus#(s(x),s(y)) -> not#(gt(x,y))
p9: plus#(s(x),s(y)) -> id#(x)
p10: plus#(s(x),s(y)) -> id#(y)
p11: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))
p12: plus#(s(x),x) -> if#(gt(x,x),id(x),id(x))
p13: plus#(s(x),x) -> gt#(x,x)
p14: plus#(s(x),x) -> id#(x)
p15: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p16: plus#(id(x),s(y)) -> if#(gt(s(y),y),y,s(y))
p17: plus#(id(x),s(y)) -> gt#(s(y),y)
p18: not#(x) -> if#(x,false(),true())
p19: 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: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}
  {p4, p11, p15}
  {p19}


-- 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: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r1, r2

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) = max{0, x1 - 5, x2 - 9}
            s_A(x1) = x1 + 4
            minus_A(x1,x2) = max{x1 + 1, x2 - 5}
            |0|_A = 2
            quot#!_A(x1,x2) = max{1, x1 - 5, x2 - 7}
            s!_A(x1) = 28
            minus!_A(x1,x2) = 0
            |0|!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x1 - 23, x2 - 45}
            s_A(x1) = x1 + 22
            minus_A(x1,x2) = max{x1 + 1, x2 - 23}
            |0|_A = 20
            quot#!_A(x1,x2) = max{4, x1 - 23, x2 - 68}
            s!_A(x1) = 3
            minus!_A(x1,x2) = max{2, x1 - 23, x2 - 47}
            |0|!_A = 3
        
  
    partial status:
  
      pi(quot#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []

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)

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: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: 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) = max{0, x1 - 74, x2 - 74}
            s_A(x1) = x1 + 73
            minus#!_A(x1,x2) = max{6, x1 - 74, x2 - 66}
            s!_A(x1) = x1 + 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{0, x1 - 23, x2 - 23}
            s_A(x1) = max{25, x1 + 22}
            minus#!_A(x1,x2) = max{3, x1 - 26, x2 - 103}
            s!_A(x1) = max{2, x1 - 104}
        
  
    partial status:
  
      pi(minus#) = []
      pi(s) = []

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: 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: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r9, r10, r11, r12, r13, 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:
          
            plus#_A(x1,x2) = max{0, x1 - 59, x2 - 59}
            s_A(x1) = max{58, x1 + 17}
            if_A(x1,x2,x3) = max{43, x1 - 60, x2, x3}
            gt_A(x1,x2) = max{18, x1 - 59, x2 - 112}
            not_A(x1) = max{103, x1 - 60}
            id_A(x1) = max{30, x1}
            true_A = 18
            false_A = 1
            zero_A = 58
            plus#!_A(x1,x2) = max{1, x1 - 56, x2 - 70}
            s!_A(x1) = max{165, x1 + 162}
            if!_A(x1,x2,x3) = max{166, x2 - 36}
            gt!_A(x1,x2) = max{3, x1 - 71, x2 - 71}
            not!_A(x1) = max{167, x1 - 72}
            id!_A(x1) = max{165, x1 - 132}
            true!_A = 163
            false!_A = 4
            zero!_A = 164
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 13, x2 - 14}
            s_A(x1) = x1 + 8
            if_A(x1,x2,x3) = max{x1 - 10, x2, x3}
            gt_A(x1,x2) = max{1, x1 - 17, x2 - 9}
            not_A(x1) = max{20, x1 - 10}
            id_A(x1) = max{12, x1 + 2}
            true_A = 1
            false_A = 1
            zero_A = 0
            plus#!_A(x1,x2) = max{14, x1 - 95, x2 - 49}
            s!_A(x1) = 8
            if!_A(x1,x2,x3) = max{12, x1 - 21, x2 - 3, x3 - 100}
            gt!_A(x1,x2) = max{2, x1 - 9, x2 - 9}
            not!_A(x1) = max{13, x1 - 20}
            id!_A(x1) = max{15, x1 - 19}
            true!_A = 1
            false!_A = 3
            zero!_A = 4
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = [1, 2, 3]
      pi(gt) = [1]
      pi(not) = [1]
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p1, p3



-- SCC decomposition.

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

p1: 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: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r10, r11, r13, 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:
          
            plus#_A(x1,x2) = max{0, x1 - 211, x2 - 129}
            id_A(x1) = max{79, x1 + 33}
            s_A(x1) = max{41, x1 + 34}
            if_A(x1,x2,x3) = max{126, x1 - 1, x2 + 34, x3}
            gt_A(x1,x2) = max{129, x1 - 42, x2 - 35}
            true_A = 129
            false_A = 3
            zero_A = 34
            plus#!_A(x1,x2) = max{36, x1 - 163, x2 - 211}
            id!_A(x1) = max{1, x1 - 171}
            s!_A(x1) = x1 + 35
            if!_A(x1,x2,x3) = max{38, x1 - 130, x2 - 80, x3 - 207}
            gt!_A(x1,x2) = max{2, x1 - 175, x2 - 176}
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 59, x2 - 20}
            id_A(x1) = max{19, x1 + 18}
            s_A(x1) = max{19, x1}
            if_A(x1,x2,x3) = max{19, x1 - 13, x2, x3}
            gt_A(x1,x2) = max{32, x1 - 25, x2 - 10}
            true_A = 19
            false_A = 32
            zero_A = 9
            plus#!_A(x1,x2) = max{1, x1 - 21, x2 - 79}
            id!_A(x1) = max{0, x1 - 22}
            s!_A(x1) = 0
            if!_A(x1,x2,x3) = max{2, x1 - 53, x2 - 43, x3 - 20}
            gt!_A(x1,x2) = max{0, x1 - 38, x2 - 23}
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(id) = [1]
      pi(s) = [1]
      pi(if) = [2, 3]
      pi(gt) = [1, 2]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

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: 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: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: 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 - 30, x2 - 30}
            s_A(x1) = max{33, x1 + 29}
            gt#!_A(x1,x2) = max{2, x1 - 30, x2 - 34}
            s!_A(x1) = max{1, x1 - 185}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = max{0, x1 - 28, x2 - 28}
            s_A(x1) = max{30, x1 + 27}
            gt#!_A(x1,x2) = max{1, x1 - 31, x2 - 31}
            s!_A(x1) = max{2, x1 - 126}
        
  
    partial status:
  
      pi(gt#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.