YES
We show the termination of the TRS R:

  min(x,|0|()) -> |0|()
  min(|0|(),y) -> |0|()
  min(s(x),s(y)) -> s(min(x,y))
  max(x,|0|()) -> x
  max(|0|(),y) -> y
  max(s(x),s(y)) -> s(max(x,y))
  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> s(minus(x,any(y)))
  gcd(s(x),s(y)) -> gcd(minus(max(x,y),min(x,y)),s(min(x,y)))
  any(s(x)) -> s(s(any(x)))
  any(x) -> x


-- SCC decomposition.

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

p1: min#(s(x),s(y)) -> min#(x,y)
p2: max#(s(x),s(y)) -> max#(x,y)
p3: minus#(s(x),s(y)) -> minus#(x,any(y))
p4: minus#(s(x),s(y)) -> any#(y)
p5: gcd#(s(x),s(y)) -> gcd#(minus(max(x,y),min(x,y)),s(min(x,y)))
p6: gcd#(s(x),s(y)) -> minus#(max(x,y),min(x,y))
p7: gcd#(s(x),s(y)) -> max#(x,y)
p8: gcd#(s(x),s(y)) -> min#(x,y)
p9: any#(s(x)) -> any#(x)

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: minus(x,|0|()) -> x
r8: minus(s(x),s(y)) -> s(minus(x,any(y)))
r9: gcd(s(x),s(y)) -> gcd(minus(max(x,y),min(x,y)),s(min(x,y)))
r10: any(s(x)) -> s(s(any(x)))
r11: any(x) -> x

The estimated dependency graph contains the following SCCs:

  {p5}
  {p1}
  {p2}
  {p3}
  {p9}


-- Reduction pair.

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

p1: gcd#(s(x),s(y)) -> gcd#(minus(max(x,y),min(x,y)),s(min(x,y)))

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: minus(x,|0|()) -> x
r8: minus(s(x),s(y)) -> s(minus(x,any(y)))
r9: gcd(s(x),s(y)) -> gcd(minus(max(x,y),min(x,y)),s(min(x,y)))
r10: any(s(x)) -> s(s(any(x)))
r11: any(x) -> x

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

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:
          
            gcd#_A(x1,x2) = max{0, x1 - 87, x2 - 136}
            s_A(x1) = max{112, x1 + 86}
            minus_A(x1,x2) = max{40, x1 + 25}
            max_A(x1,x2) = max{39, x1 + 13, x2 + 3}
            min_A(x1,x2) = max{51, x1 + 25}
            |0|_A = 27
            any_A(x1) = max{1, x1 - 113}
            gcd#!_A(x1,x2) = max{24, x1 - 39, x2 - 82}
            s!_A(x1) = x1 + 6
            minus!_A(x1,x2) = max{5, x1 - 32}
            max!_A(x1,x2) = max{46, x1 - 28, x2 - 68}
            min!_A(x1,x2) = max{3, x1 - 28}
            |0|!_A = 1
            any!_A(x1) = max{4, x1 - 108}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              gcd#_A(x1,x2) = x2
              s_A(x1) = x1 + 2
              minus_A(x1,x2) = x1 + 3
              max_A(x1,x2) = x1 + x2 + 1
              min_A(x1,x2) = x2
              |0|_A() = 0
              any_A(x1) = 2
              gcd#!_A(x1,x2) = 3
              s!_A(x1) = x1 + 2
              minus!_A(x1,x2) = 4
              max!_A(x1,x2) = x1 + x2 + 4
              min!_A(x1,x2) = x1 + 1
              |0|!_A() = 0
              any!_A(x1) = 5
        
  
    partial status:
  
      pi(gcd#) = [1, 2]
      pi(s) = [1]
      pi(minus) = [1]
      pi(max) = [1, 2]
      pi(min) = []
      pi(|0|) = []
      pi(any) = [1]

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

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: minus(x,|0|()) -> x
r8: minus(s(x),s(y)) -> s(minus(x,any(y)))
r9: gcd(s(x),s(y)) -> gcd(minus(max(x,y),min(x,y)),s(min(x,y)))
r10: any(s(x)) -> s(s(any(x)))
r11: any(x) -> x

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:
          
            min#_A(x1,x2) = 0
            s_A(x1) = max{7, x1 + 3}
            min_A(x1,x2) = max{29, x1 - 8, x2 - 8}
            |0|_A = 7
            min#!_A(x1,x2) = max{x1 - 1, x2 + 9}
            s!_A(x1) = 0
            min!_A(x1,x2) = max{1, x1 - 39, x2 - 31}
            |0|!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              min#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 3
              min_A(x1,x2) = x1 + x2 + 1
              |0|_A() = 1
              min#!_A(x1,x2) = x2
              s!_A(x1) = 8
              min!_A(x1,x2) = 1
              |0|!_A() = 0
        
  
    partial status:
  
      pi(min#) = [2]
      pi(s) = [1]
      pi(min) = [1, 2]
      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: max#(s(x),s(y)) -> max#(x,y)

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: minus(x,|0|()) -> x
r8: minus(s(x),s(y)) -> s(minus(x,any(y)))
r9: gcd(s(x),s(y)) -> gcd(minus(max(x,y),min(x,y)),s(min(x,y)))
r10: any(s(x)) -> s(s(any(x)))
r11: any(x) -> x

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:
          
            max#_A(x1,x2) = 0
            s_A(x1) = max{6, x1 - 8}
            max#!_A(x1,x2) = 52
            s!_A(x1) = x1 + 51
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              max#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              max#!_A(x1,x2) = x1 + x2
              s!_A(x1) = 3
        
  
    partial status:
  
      pi(max#) = []
      pi(s) = [1]

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,any(y))

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: minus(x,|0|()) -> x
r8: minus(s(x),s(y)) -> s(minus(x,any(y)))
r9: gcd(s(x),s(y)) -> gcd(minus(max(x,y),min(x,y)),s(min(x,y)))
r10: any(s(x)) -> s(s(any(x)))
r11: any(x) -> x

The set of usable rules consists of

  r11

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{14, x1 - 4}
            any_A(x1) = max{9, x1 + 4}
            minus#!_A(x1,x2) = 0
            s!_A(x1) = max{0, x1 - 57}
            any!_A(x1) = max{0, x1 - 43}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              minus#_A(x1,x2) = x1
              s_A(x1) = x1
              any_A(x1) = x1
              minus#!_A(x1,x2) = 1
              s!_A(x1) = x1 + 1
              any!_A(x1) = x1 + 2
        
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [1]
      pi(any) = [1]

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: any#(s(x)) -> any#(x)

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: minus(x,|0|()) -> x
r8: minus(s(x),s(y)) -> s(minus(x,any(y)))
r9: gcd(s(x),s(y)) -> gcd(minus(max(x,y),min(x,y)),s(min(x,y)))
r10: any(s(x)) -> s(s(any(x)))
r11: any(x) -> x

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:
          
            any#_A(x1) = 0
            s_A(x1) = max{2, x1 - 10}
            min_A(x1,x2) = max{x1, x2}
            |0|_A = 1
            any#!_A(x1) = 1
            s!_A(x1) = 3
            min!_A(x1,x2) = 2
            |0|!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              any#_A(x1) = x1
              s_A(x1) = x1 + 2
              min_A(x1,x2) = x1 + x2 + 1
              |0|_A() = 1
              any#!_A(x1) = x1
              s!_A(x1) = x1 + 4
              min!_A(x1,x2) = 3
              |0|!_A() = 0
        
  
    partial status:
  
      pi(any#) = [1]
      pi(s) = [1]
      pi(min) = [1]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.