YES

We show the termination of the TRS R:

  t(N) -> cs(r(q(N)),nt(ns(N)))
  q(|0|()) -> |0|()
  q(s(X)) -> s(p(q(X),d(X)))
  d(|0|()) -> |0|()
  d(s(X)) -> s(s(d(X)))
  p(|0|(),X) -> X
  p(X,|0|()) -> X
  p(s(X),s(Y)) -> s(s(p(X,Y)))
  f(|0|(),X) -> nil()
  f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
  t(X) -> nt(X)
  s(X) -> ns(X)
  f(X1,X2) -> nf(X1,X2)
  a(nt(X)) -> t(a(X))
  a(ns(X)) -> s(a(X))
  a(nf(X1,X2)) -> f(a(X1),a(X2))
  a(X) -> X

-- SCC decomposition.

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

p1: t#(N) -> q#(N)
p2: q#(s(X)) -> s#(p(q(X),d(X)))
p3: q#(s(X)) -> p#(q(X),d(X))
p4: q#(s(X)) -> q#(X)
p5: q#(s(X)) -> d#(X)
p6: d#(s(X)) -> s#(s(d(X)))
p7: d#(s(X)) -> s#(d(X))
p8: d#(s(X)) -> d#(X)
p9: p#(s(X),s(Y)) -> s#(s(p(X,Y)))
p10: p#(s(X),s(Y)) -> s#(p(X,Y))
p11: p#(s(X),s(Y)) -> p#(X,Y)
p12: f#(s(X),cs(Y,Z)) -> a#(Z)
p13: a#(nt(X)) -> t#(a(X))
p14: a#(nt(X)) -> a#(X)
p15: a#(ns(X)) -> s#(a(X))
p16: a#(ns(X)) -> a#(X)
p17: a#(nf(X1,X2)) -> f#(a(X1),a(X2))
p18: a#(nf(X1,X2)) -> a#(X1)
p19: a#(nf(X1,X2)) -> a#(X2)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The estimated dependency graph contains the following SCCs:

  {p12, p14, p16, p17, p18, p19}
  {p4}
  {p11}
  {p8}


-- Reduction pair.

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

p1: a#(nf(X1,X2)) -> a#(X2)
p2: a#(nf(X1,X2)) -> a#(X1)
p3: a#(nf(X1,X2)) -> f#(a(X1),a(X2))
p4: f#(s(X),cs(Y,Z)) -> a#(Z)
p5: a#(ns(X)) -> a#(X)
p6: a#(nt(X)) -> a#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = ((0,0),(1,0)) x1 + (1,4)
            nf_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,5)
            f#_A(x1,x2) = ((0,0),(1,0)) x2 + (1,6)
            a_A(x1) = x1 + (0,8)
            s_A(x1) = ((1,0),(0,0)) x1 + (0,10)
            cs_A(x1,x2) = ((1,0),(0,0)) x2 + (0,11)
            ns_A(x1) = ((1,0),(0,0)) x1 + (0,9)
            nt_A(x1) = ((1,0),(0,0)) x1 + (0,9)
            d_A(x1) = (0,11)
            |0|_A() = (0,11)
            p_A(x1,x2) = x1 + x2 + (0,11)
            q_A(x1) = (1,12)
            t_A(x1) = ((1,0),(0,0)) x1 + (0,17)
            r_A(x1) = (1,1)
            f_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,12)
            nil_A() = (0,11)
    
      precedence:
      
        p = f > a = t = r > a# = f# = cs > nil > nt = q > nf = s = ns = d = |0|
      
      partial status:
    
        pi(a#) = []
        pi(nf) = []
        pi(f#) = []
        pi(a) = [1]
        pi(s) = []
        pi(cs) = []
        pi(ns) = []
        pi(nt) = []
        pi(d) = []
        pi(|0|) = []
        pi(p) = [1, 2]
        pi(q) = []
        pi(t) = []
        pi(r) = []
        pi(f) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = (1,1)
            nf_A(x1,x2) = (2,5)
            f#_A(x1,x2) = (1,1)
            a_A(x1) = ((1,0),(1,1)) x1 + (0,3)
            s_A(x1) = (3,8)
            cs_A(x1,x2) = (4,9)
            ns_A(x1) = (0,4)
            nt_A(x1) = (5,2)
            d_A(x1) = (3,9)
            |0|_A() = (1,0)
            p_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (4,9)
            q_A(x1) = (4,2)
            t_A(x1) = (5,3)
            r_A(x1) = (0,0)
            f_A(x1,x2) = (1,4)
            nil_A() = (0,0)
    
      precedence:
      
        a = cs = d > s = |0| = p = q > ns = t = r > f = nil > a# = nf = f# = nt
      
      partial status:
    
        pi(a#) = []
        pi(nf) = []
        pi(f#) = []
        pi(a) = []
        pi(s) = []
        pi(cs) = []
        pi(ns) = []
        pi(nt) = []
        pi(d) = []
        pi(|0|) = []
        pi(p) = []
        pi(q) = []
        pi(t) = []
        pi(r) = []
        pi(f) = []
        pi(nil) = []
    

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: a#(nf(X1,X2)) -> a#(X2)
p2: a#(nf(X1,X2)) -> f#(a(X1),a(X2))
p3: f#(s(X),cs(Y,Z)) -> a#(Z)
p4: a#(ns(X)) -> a#(X)
p5: a#(nt(X)) -> a#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

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: a#(nf(X1,X2)) -> a#(X2)
p2: a#(nt(X)) -> a#(X)
p3: a#(ns(X)) -> a#(X)
p4: a#(nf(X1,X2)) -> f#(a(X1),a(X2))
p5: f#(s(X),cs(Y,Z)) -> a#(Z)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = ((0,0),(1,0)) x1 + (3,0)
            nf_A(x1,x2) = ((1,0),(1,0)) x2 + (0,2)
            nt_A(x1) = ((1,0),(0,0)) x1 + (2,1)
            ns_A(x1) = x1
            f#_A(x1,x2) = ((0,0),(1,0)) x2 + (3,0)
            a_A(x1) = x1 + (0,3)
            s_A(x1) = x1
            cs_A(x1,x2) = ((1,0),(0,0)) x2
            d_A(x1) = (0,0)
            |0|_A() = (0,0)
            p_A(x1,x2) = x1 + x2
            q_A(x1) = (1,1)
            t_A(x1) = ((1,0),(0,0)) x1 + (2,2)
            r_A(x1) = (1,1)
            f_A(x1,x2) = ((1,0),(1,0)) x2 + (0,4)
            nil_A() = (0,0)
    
      precedence:
      
        a = p > a# = f# > t = f > nf > nt > d = q = nil > s > ns = r > cs = |0|
      
      partial status:
    
        pi(a#) = []
        pi(nf) = []
        pi(nt) = []
        pi(ns) = []
        pi(f#) = []
        pi(a) = []
        pi(s) = []
        pi(cs) = []
        pi(d) = []
        pi(|0|) = []
        pi(p) = [1, 2]
        pi(q) = []
        pi(t) = []
        pi(r) = []
        pi(f) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = (0,2)
            nf_A(x1,x2) = (0,3)
            nt_A(x1) = (1,1)
            ns_A(x1) = (0,0)
            f#_A(x1,x2) = (0,2)
            a_A(x1) = (3,6)
            s_A(x1) = (2,1)
            cs_A(x1,x2) = (0,4)
            d_A(x1) = (3,2)
            |0|_A() = (1,3)
            p_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (3,2)
            q_A(x1) = (3,8)
            t_A(x1) = (2,7)
            r_A(x1) = (0,0)
            f_A(x1,x2) = (0,5)
            nil_A() = (0,0)
    
      precedence:
      
        p > q > f > |0| > nt = t > r > a = cs > s > a# = nf = f# > ns = d = nil
      
      partial status:
    
        pi(a#) = []
        pi(nf) = []
        pi(nt) = []
        pi(ns) = []
        pi(f#) = []
        pi(a) = []
        pi(s) = []
        pi(cs) = []
        pi(d) = []
        pi(|0|) = []
        pi(p) = [1]
        pi(q) = []
        pi(t) = []
        pi(r) = []
        pi(f) = []
        pi(nil) = []
    

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: a#(nf(X1,X2)) -> a#(X2)
p2: a#(ns(X)) -> a#(X)
p3: a#(nf(X1,X2)) -> f#(a(X1),a(X2))
p4: f#(s(X),cs(Y,Z)) -> a#(Z)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

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: a#(nf(X1,X2)) -> a#(X2)
p2: a#(nf(X1,X2)) -> f#(a(X1),a(X2))
p3: f#(s(X),cs(Y,Z)) -> a#(Z)
p4: a#(ns(X)) -> a#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = ((1,0),(1,0)) x1 + (0,3)
            nf_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (6,4)
            f#_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (5,1)
            a_A(x1) = ((1,0),(1,1)) x1 + (0,7)
            s_A(x1) = ((1,0),(1,0)) x1 + (0,8)
            cs_A(x1,x2) = ((1,0),(0,0)) x2 + (0,5)
            ns_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            d_A(x1) = (0,9)
            |0|_A() = (0,1)
            p_A(x1,x2) = x1 + x2 + (0,9)
            q_A(x1) = (1,10)
            t_A(x1) = (5,11)
            r_A(x1) = (1,1)
            nt_A(x1) = (5,0)
            f_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (6,6)
            nil_A() = (0,0)
    
      precedence:
      
        a# = nf = f# = a = s = cs = ns = d = |0| = p = q = t = r = nt = f = nil
      
      partial status:
    
        pi(a#) = []
        pi(nf) = []
        pi(f#) = []
        pi(a) = []
        pi(s) = []
        pi(cs) = []
        pi(ns) = []
        pi(d) = []
        pi(|0|) = []
        pi(p) = []
        pi(q) = []
        pi(t) = []
        pi(r) = []
        pi(nt) = []
        pi(f) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = (6,9)
            nf_A(x1,x2) = (0,0)
            f#_A(x1,x2) = ((1,0),(1,1)) x1 + (2,1)
            a_A(x1) = (3,0)
            s_A(x1) = (5,8)
            cs_A(x1,x2) = (1,0)
            ns_A(x1) = (0,0)
            d_A(x1) = (7,0)
            |0|_A() = (6,0)
            p_A(x1,x2) = x1 + ((1,0),(1,1)) x2
            q_A(x1) = (6,0)
            t_A(x1) = (3,0)
            r_A(x1) = (0,0)
            nt_A(x1) = (1,1)
            f_A(x1,x2) = (2,1)
            nil_A() = (0,0)
    
      precedence:
      
        a > p > t = r > nt > d = q > s > ns > a# = f# = cs > nf = |0| = f = nil
      
      partial status:
    
        pi(a#) = []
        pi(nf) = []
        pi(f#) = []
        pi(a) = []
        pi(s) = []
        pi(cs) = []
        pi(ns) = []
        pi(d) = []
        pi(|0|) = []
        pi(p) = []
        pi(q) = []
        pi(t) = []
        pi(r) = []
        pi(nt) = []
        pi(f) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(nf(X1,X2)) -> f#(a(X1),a(X2))
p2: f#(s(X),cs(Y,Z)) -> a#(Z)
p3: a#(ns(X)) -> a#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

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: a#(nf(X1,X2)) -> f#(a(X1),a(X2))
p2: f#(s(X),cs(Y,Z)) -> a#(Z)
p3: a#(ns(X)) -> a#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = ((1,0),(0,0)) x1 + (0,8)
            nf_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,10)
            f#_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,1)
            a_A(x1) = x1 + (0,3)
            s_A(x1) = ((1,0),(0,0)) x1 + (0,12)
            cs_A(x1,x2) = ((1,0),(0,0)) x2 + (0,9)
            ns_A(x1) = ((1,0),(0,0)) x1 + (0,9)
            d_A(x1) = (0,13)
            |0|_A() = (0,1)
            p_A(x1,x2) = x1 + x2 + (0,13)
            q_A(x1) = x1 + (1,11)
            t_A(x1) = (3,10)
            r_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            nt_A(x1) = (3,9)
            f_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,10)
            nil_A() = (0,0)
    
      precedence:
      
        a# = a > r > f > nf > d > |0| = nil > cs = p = q = t > f# > s = nt > ns
      
      partial status:
    
        pi(a#) = []
        pi(nf) = []
        pi(f#) = [1, 2]
        pi(a) = [1]
        pi(s) = []
        pi(cs) = []
        pi(ns) = []
        pi(d) = []
        pi(|0|) = []
        pi(p) = [1, 2]
        pi(q) = [1]
        pi(t) = []
        pi(r) = []
        pi(nt) = []
        pi(f) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = (10,10)
            nf_A(x1,x2) = (15,13)
            f#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (6,1)
            a_A(x1) = ((1,0),(1,0)) x1 + (2,2)
            s_A(x1) = (14,12)
            cs_A(x1,x2) = (11,11)
            ns_A(x1) = (13,1)
            d_A(x1) = (15,0)
            |0|_A() = (0,0)
            p_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (15,1)
            q_A(x1) = ((0,0),(1,0)) x1 + (16,0)
            t_A(x1) = (12,12)
            r_A(x1) = (0,0)
            nt_A(x1) = (11,1)
            f_A(x1,x2) = (16,14)
            nil_A() = (0,0)
    
      precedence:
      
        a = p = nt > t > a# = nf = f# = s = d = |0| = q = nil > r = f > cs = ns
      
      partial status:
    
        pi(a#) = []
        pi(nf) = []
        pi(f#) = [1, 2]
        pi(a) = []
        pi(s) = []
        pi(cs) = []
        pi(ns) = []
        pi(d) = []
        pi(|0|) = []
        pi(p) = [1]
        pi(q) = []
        pi(t) = []
        pi(r) = []
        pi(nt) = []
        pi(f) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(ns(X)) -> a#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: a#(ns(X)) -> a#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            ns_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        a# = ns
      
      partial status:
    
        pi(a#) = []
        pi(ns) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = ((1,0),(0,0)) x1
            ns_A(x1) = (1,1)
    
      precedence:
      
        ns > a#
      
      partial status:
    
        pi(a#) = []
        pi(ns) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: q#(s(X)) -> q#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            q#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        q# = s
      
      partial status:
    
        pi(q#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            q#_A(x1) = ((1,0),(0,0)) x1
            s_A(x1) = (1,1)
    
      precedence:
      
        s > q#
      
      partial status:
    
        pi(q#) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: p#(s(X),s(Y)) -> p#(X,Y)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            p#_A(x1,x2) = ((1,0),(0,0)) x1
            s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        p# > s
      
      partial status:
    
        pi(p#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            p#_A(x1,x2) = (0,0)
            s_A(x1) = (1,1)
    
      precedence:
      
        p# = s
      
      partial status:
    
        pi(p#) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: d#(s(X)) -> d#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            d#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        d# = s
      
      partial status:
    
        pi(d#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            d#_A(x1) = ((1,0),(0,0)) x1
            s_A(x1) = (1,1)
    
      precedence:
      
        s > d#
      
      partial status:
    
        pi(d#) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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