YES

We show the termination of the relative TRS R/S:

  R:
  eq(|0|(),|0|()) -> true()
  eq(|0|(),s(x)) -> false()
  eq(s(x),|0|()) -> false()
  eq(s(x),s(y)) -> eq(x,y)
  le(|0|(),y) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  app(nil(),y) -> y
  app(add(n,x),y) -> add(n,app(x,y))
  min(add(n,nil())) -> n
  min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
  if_min(true(),add(n,add(m,x))) -> min(add(n,x))
  if_min(false(),add(n,add(m,x))) -> min(add(m,x))
  rm(n,nil()) -> nil()
  rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
  if_rm(true(),n,add(m,x)) -> rm(n,x)
  if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
  minsort(nil(),nil()) -> nil()
  minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
  if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
  if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

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

p1: eq#(s(x),s(y)) -> eq#(x,y)
p2: le#(s(x),s(y)) -> le#(x,y)
p3: app#(add(n,x),y) -> app#(x,y)
p4: min#(add(n,add(m,x))) -> if_min#(le(n,m),add(n,add(m,x)))
p5: min#(add(n,add(m,x))) -> le#(n,m)
p6: if_min#(true(),add(n,add(m,x))) -> min#(add(n,x))
p7: if_min#(false(),add(n,add(m,x))) -> min#(add(m,x))
p8: rm#(n,add(m,x)) -> if_rm#(eq(n,m),n,add(m,x))
p9: rm#(n,add(m,x)) -> eq#(n,m)
p10: if_rm#(true(),n,add(m,x)) -> rm#(n,x)
p11: if_rm#(false(),n,add(m,x)) -> rm#(n,x)
p12: minsort#(add(n,x),y) -> if_minsort#(eq(n,min(add(n,x))),add(n,x),y)
p13: minsort#(add(n,x),y) -> eq#(n,min(add(n,x)))
p14: minsort#(add(n,x),y) -> min#(add(n,x))
p15: if_minsort#(true(),add(n,x),y) -> minsort#(app(rm(n,x),y),nil())
p16: if_minsort#(true(),add(n,x),y) -> app#(rm(n,x),y)
p17: if_minsort#(true(),add(n,x),y) -> rm#(n,x)
p18: if_minsort#(false(),add(n,x),y) -> minsort#(x,add(n,y))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))
r22: rand(x) -> x
r23: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p12, p15, p18}
  {p8, p10, p11}
  {p1}
  {p4, p6, p7}
  {p2}
  {p3}


-- Reduction pair.

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

p1: if_minsort#(false(),add(n,x),y) -> minsort#(x,add(n,y))
p2: minsort#(add(n,x),y) -> if_minsort#(eq(n,min(add(n,x))),add(n,x),y)
p3: if_minsort#(true(),add(n,x),y) -> minsort#(app(rm(n,x),y),nil())

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))
r22: rand(x) -> x
r23: rand(x) -> rand(s(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, r18, r19, r20, r21, r22, r23

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        if_minsort#_A(x1,x2,x3) = ((1,0),(1,1)) x2 + x3
        false_A() = (1,4)
        add_A(x1,x2) = x1 + x2 + (3,1)
        minsort#_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (0,3)
        eq_A(x1,x2) = (2,5)
        min_A(x1) = x1 + (1,5)
        true_A() = (2,2)
        app_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (0,2)
        rm_A(x1,x2) = x2 + (0,3)
        nil_A() = (0,0)
        |0|_A() = (1,1)
        s_A(x1) = x1
        le_A(x1,x2) = ((1,0),(1,1)) x2 + (3,1)
        if_min_A(x1,x2) = ((1,0),(1,1)) x2
        if_rm_A(x1,x2,x3) = x3 + (0,2)
        minsort_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (1,2)
        if_minsort_A(x1,x2,x3) = ((1,0),(1,1)) x2 + x3 + (1,0)
        rand_A(x1) = x1 + (1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > if_rm > if_min > app > minsort# > eq > add > true > if_minsort# > false > minsort > rm > if_minsort > min > le > |0| > nil
      
      argument filter:
    
        pi(if_minsort#) = [2]
        pi(false) = []
        pi(add) = []
        pi(minsort#) = []
        pi(eq) = []
        pi(min) = 1
        pi(true) = []
        pi(app) = [2]
        pi(rm) = []
        pi(nil) = []
        pi(|0|) = []
        pi(s) = 1
        pi(le) = 2
        pi(if_min) = 2
        pi(if_rm) = []
        pi(minsort) = []
        pi(if_minsort) = [2, 3]
        pi(rand) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3

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

-- Reduction pair.

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

p1: if_rm#(false(),n,add(m,x)) -> rm#(n,x)
p2: rm#(n,add(m,x)) -> if_rm#(eq(n,m),n,add(m,x))
p3: if_rm#(true(),n,add(m,x)) -> rm#(n,x)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))
r22: rand(x) -> x
r23: rand(x) -> rand(s(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, r18, r19, r20, r21, r22, r23

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        if_rm#_A(x1,x2,x3) = x2 + x3 + (0,4)
        false_A() = (1,3)
        add_A(x1,x2) = x1 + x2 + (2,1)
        rm#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,0)
        eq_A(x1,x2) = x1 + x2 + (1,1)
        true_A() = (1,1)
        |0|_A() = (1,1)
        s_A(x1) = ((1,0),(1,1)) x1 + (0,3)
        le_A(x1,x2) = x2 + (2,3)
        app_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2
        nil_A() = (0,0)
        min_A(x1) = x1 + (1,1)
        if_min_A(x1,x2) = x2
        rm_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (0,3)
        if_rm_A(x1,x2,x3) = x3 + (0,2)
        minsort_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (1,1)
        if_minsort_A(x1,x2,x3) = ((1,0),(1,0)) x2 + x3 + (1,0)
        rand_A(x1) = x1 + (1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        eq > false > if_rm# > rand > if_minsort > minsort > true > rm > min > app > add > if_rm > if_min > |0| > nil > le > s > rm#
      
      argument filter:
    
        pi(if_rm#) = []
        pi(false) = []
        pi(add) = []
        pi(rm#) = []
        pi(eq) = [1]
        pi(true) = []
        pi(|0|) = []
        pi(s) = 1
        pi(le) = 2
        pi(app) = [2]
        pi(nil) = []
        pi(min) = []
        pi(if_min) = []
        pi(rm) = []
        pi(if_rm) = []
        pi(minsort) = []
        pi(if_minsort) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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

-- Reduction pair.

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

p1: eq#(s(x),s(y)) -> eq#(x,y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))
r22: rand(x) -> x
r23: rand(x) -> rand(s(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, r18, r19, r20, r21, r22, r23

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        eq#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2
        s_A(x1) = x1 + (0,1)
        eq_A(x1,x2) = (2,5)
        |0|_A() = (1,1)
        true_A() = (1,3)
        false_A() = (1,4)
        le_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (2,1)
        app_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (0,1)
        nil_A() = (0,1)
        add_A(x1,x2) = x1 + x2 + (2,2)
        min_A(x1) = x1 + (1,3)
        if_min_A(x1,x2) = ((1,0),(1,1)) x2
        rm_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (0,2)
        if_rm_A(x1,x2,x3) = ((0,0),(1,0)) x2 + x3 + (0,3)
        minsort_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (1,2)
        if_minsort_A(x1,x2,x3) = ((1,0),(1,0)) x2 + x3 + (1,1)
        rand_A(x1) = x1 + (1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > if_minsort > eq > true > minsort > false > app > add > rm > if_rm > if_min > le > min > nil > |0| > eq#
      
      argument filter:
    
        pi(eq#) = [1]
        pi(s) = [1]
        pi(eq) = []
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(le) = [1]
        pi(app) = []
        pi(nil) = []
        pi(add) = []
        pi(min) = 1
        pi(if_min) = [2]
        pi(rm) = []
        pi(if_rm) = []
        pi(minsort) = []
        pi(if_minsort) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: if_min#(false(),add(n,add(m,x))) -> min#(add(m,x))
p2: min#(add(n,add(m,x))) -> if_min#(le(n,m),add(n,add(m,x)))
p3: if_min#(true(),add(n,add(m,x))) -> min#(add(n,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))
r22: rand(x) -> x
r23: rand(x) -> rand(s(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, r18, r19, r20, r21, r22, r23

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        if_min#_A(x1,x2) = ((1,0),(1,1)) x2
        false_A() = (1,0)
        add_A(x1,x2) = x1 + x2 + (4,3)
        min#_A(x1) = ((1,0),(1,0)) x1 + (1,6)
        le_A(x1,x2) = x1 + (2,2)
        true_A() = (1,1)
        eq_A(x1,x2) = (1,2)
        |0|_A() = (1,0)
        s_A(x1) = ((1,0),(1,1)) x1 + (0,1)
        app_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (0,1)
        nil_A() = (0,7)
        min_A(x1) = x1 + (1,1)
        if_min_A(x1,x2) = x2
        rm_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (0,5)
        if_rm_A(x1,x2,x3) = ((1,0),(1,0)) x3
        minsort_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (1,6)
        if_minsort_A(x1,x2,x3) = x1 + ((1,0),(1,1)) x2 + x3
        rand_A(x1) = ((1,0),(1,0)) x1 + (1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        rand > minsort > s > eq > true > false > rm > if_rm > le > add > if_minsort > min > if_min > nil > app > |0| > min# > if_min#
      
      argument filter:
    
        pi(if_min#) = 2
        pi(false) = []
        pi(add) = []
        pi(min#) = []
        pi(le) = 1
        pi(true) = []
        pi(eq) = []
        pi(|0|) = []
        pi(s) = []
        pi(app) = [1]
        pi(nil) = []
        pi(min) = []
        pi(if_min) = []
        pi(rm) = []
        pi(if_rm) = []
        pi(minsort) = []
        pi(if_minsort) = [1]
        pi(rand) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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

-- Reduction pair.

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

p1: le#(s(x),s(y)) -> le#(x,y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))
r22: rand(x) -> x
r23: rand(x) -> rand(s(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, r18, r19, r20, r21, r22, r23

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        le#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2
        s_A(x1) = x1 + (0,1)
        eq_A(x1,x2) = x1 + x2 + (1,3)
        |0|_A() = (1,1)
        true_A() = (1,1)
        false_A() = (1,4)
        le_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (2,1)
        app_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (0,1)
        nil_A() = (0,0)
        add_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (2,1)
        min_A(x1) = x1 + (1,6)
        if_min_A(x1,x2) = ((1,0),(1,1)) x2
        rm_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (0,3)
        if_rm_A(x1,x2,x3) = ((1,0),(1,0)) x3
        minsort_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (1,0)
        if_minsort_A(x1,x2,x3) = ((1,0),(1,0)) x2 + x3 + (1,0)
        rand_A(x1) = x1 + (1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        le# > s > rand > |0| > eq > false > true > minsort > if_minsort > rm > if_rm > if_min > le > min > add > nil > app
      
      argument filter:
    
        pi(le#) = [1]
        pi(s) = 1
        pi(eq) = 1
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(le) = [1]
        pi(app) = 2
        pi(nil) = []
        pi(add) = []
        pi(min) = []
        pi(if_min) = 2
        pi(rm) = []
        pi(if_rm) = []
        pi(minsort) = [2]
        pi(if_minsort) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: app#(add(n,x),y) -> app#(x,y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))
r22: rand(x) -> x
r23: rand(x) -> rand(s(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, r18, r19, r20, r21, r22, r23

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        app#_A(x1,x2) = x1
        add_A(x1,x2) = x1 + x2 + (3,1)
        eq_A(x1,x2) = ((1,0),(1,1)) x1 + (2,1)
        |0|_A() = (1,2)
        true_A() = (1,2)
        s_A(x1) = ((1,0),(1,1)) x1 + (0,1)
        false_A() = (1,3)
        le_A(x1,x2) = (2,1)
        app_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (0,1)
        nil_A() = (0,1)
        min_A(x1) = ((1,0),(1,1)) x1 + (1,2)
        if_min_A(x1,x2) = ((1,0),(1,1)) x2
        rm_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (0,4)
        if_rm_A(x1,x2,x3) = ((1,0),(1,0)) x3
        minsort_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (0,1)
        if_minsort_A(x1,x2,x3) = ((1,0),(1,0)) x2 + x3
        rand_A(x1) = ((1,0),(1,0)) x1 + (1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > if_rm > if_min > eq > false > |0| > true > add > minsort > app > if_minsort > nil > rm > le > min > app#
      
      argument filter:
    
        pi(app#) = []
        pi(add) = []
        pi(eq) = [1]
        pi(|0|) = []
        pi(true) = []
        pi(s) = []
        pi(false) = []
        pi(le) = []
        pi(app) = [1]
        pi(nil) = []
        pi(min) = 1
        pi(if_min) = 2
        pi(rm) = []
        pi(if_rm) = []
        pi(minsort) = [2]
        pi(if_minsort) = 3
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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