YES

We show the termination of the TRS R:

  plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
  plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

-- SCC decomposition.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X),plus(Y,Z)) -> plus#(s(s(Y)),Z)
p3: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))
p4: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X3,plus(X2,X4))
p5: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X2,X4)

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X2,X4)
p3: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X3,plus(X2,X4))
p4: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))
p5: plus#(s(X),plus(Y,Z)) -> plus#(s(s(Y)),Z)

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          plus#_A(x1,x2) = x2
          s_A(x1) = x1 + 2
          plus_A(x1,x2) = x2 + 3
  
    precedence:
    
      plus# = s = plus
    
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(plus) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X2,X4)
p3: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))
p4: plus#(s(X),plus(Y,Z)) -> plus#(s(s(Y)),Z)

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X),plus(Y,Z)) -> plus#(s(s(Y)),Z)
p3: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))
p4: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X2,X4)

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          plus#_A(x1,x2) = x2
          s_A(x1) = 5
          plus_A(x1,x2) = x2 + 2
  
    precedence:
    
      plus# = s = plus
    
    partial status:
  
      pi(plus#) = [2]
      pi(s) = []
      pi(plus) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X),plus(Y,Z)) -> plus#(s(s(Y)),Z)
p3: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

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: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))
p3: plus#(s(X),plus(Y,Z)) -> plus#(s(s(Y)),Z)

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          plus#_A(x1,x2) = x2 + 1
          s_A(x1) = 8
          plus_A(x1,x2) = x2 + 3
  
    precedence:
    
      plus > plus# = s
    
    partial status:
  
      pi(plus#) = [2]
      pi(s) = []
      pi(plus) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

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),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          plus#_A(x1,x2) = x1 + x2 + 1
          s_A(x1) = x1 + 3
          plus_A(x1,x2) = 1
  
    precedence:
    
      plus# = s = plus
    
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(plus) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          plus#_A(x1,x2) = x1 + x2
          s_A(x1) = x1 + 1
          plus_A(x1,x2) = 0
  
    precedence:
    
      plus# = s = plus
    
    partial status:
  
      pi(plus#) = [1, 2]
      pi(s) = []
      pi(plus) = []

The next rules are strictly ordered:

  p1

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