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^2
        order: standard order
        interpretations:
          plus#_A(x1,x2) = ((0,1),(0,0)) x2 + (3,7)
          s_A(x1) = (1,1)
          plus_A(x1,x2) = ((0,0),(1,1)) x2 + (2,2)
  
    precedence:
    
      plus# = s = plus
    
    partial status:
  
      pi(plus#) = []
      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#(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^2
        order: standard order
        interpretations:
          plus#_A(x1,x2) = ((0,1),(0,0)) x2 + (9,5)
          s_A(x1) = ((0,0),(0,1)) x1 + (8,3)
          plus_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,0),(0,1)) x2 + (1,2)
  
    precedence:
    
      plus > plus# = s
    
    partial status:
  
      pi(plus#) = []
      pi(s) = []
      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))
p2: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))
p3: 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}


-- 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#(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^2
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x2 + (1,2)
          s_A(x1) = (2,0)
          plus_A(x1,x2) = ((0,1),(1,0)) x2 + (3,1)
  
    precedence:
    
      s > plus# = plus
    
    partial status:
  
      pi(plus#) = [2]
      pi(s) = []
      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))
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^2
        order: standard order
        interpretations:
          plus#_A(x1,x2) = ((0,1),(0,1)) x1 + (3,7)
          s_A(x1) = ((0,0),(0,1)) x1 + (1,2)
          plus_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,0),(0,1)) x2 + (2,3)
  
    precedence:
    
      plus# > s = plus
    
    partial status:
  
      pi(plus#) = []
      pi(s) = []
      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^2
        order: standard order
        interpretations:
          plus#_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (0,1)
          s_A(x1) = ((1,0),(0,0)) x1 + (2,1)
          plus_A(x1,x2) = ((0,0),(1,1)) x2 + (3,2)
  
    precedence:
    
      plus# = s = plus
    
    partial status:
  
      pi(plus#) = [2]
      pi(s) = []
      pi(plus) = []

The next rules are strictly ordered:

  p1

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