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-minimaldependency 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-minimaldependency 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      if_minsort#_A(x1,x2,x3) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,0,0,0)) x2 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,0)) x3 + (0,0,0,15)
      false_A() = (1,6,0,3)
      add_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,0,0)) x2 + (9,1,9,15)
      minsort#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,0,0),(1,0,1,0)) x1 + x2 + (0,8,8,0)
      eq_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,1,1,0),(1,0,0,0)) x1 + (2,1,1,4)
      min_A(x1) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,0,19,8)
      true_A() = (0,4,1,1)
      app_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,1,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,1,1,0)) x2
      rm_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,1,0,0)) x2 + (0,0,1,3)
      nil_A() = (0,2,1,4)
      |0|_A() = (1,1,1,1)
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(1,1,1,0)) x1 + (0,1,1,1)
      le_A(x1,x2) = x1 + ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,0,0)) x2 + (1,5,2,2)
      if_min_A(x1,x2) = x2 + (0,8,0,0)
      if_rm_A(x1,x2,x3) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,1,0,0)) x3 + (0,0,2,13)
      minsort_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(0,1,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,0,0)) x2 + (1,3,0,1)
      if_minsort_A(x1,x2,x3) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(0,1,0,0)) x2 + ((1,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,0,0)) x3 + (1,3,0,0)
      rand_A(x1) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,1,1,0)

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-minimaldependency 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      if_rm#_A(x1,x2,x3) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,0,0,0)) x2 + ((1,0,0,0),(0,1,0,0),(1,1,1,0),(0,1,0,0)) x3 + (0,0,0,1)
      false_A() = (1,6,1,2)
      add_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + x2 + (3,0,4,6)
      rm#_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,1,0)) x2 + (3,6,6,0)
      eq_A(x1,x2) = (2,3,2,3)
      true_A() = (1,2,0,2)
      |0|_A() = (1,1,1,1)
      s_A(x1) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,1,0)) x1 + (0,1,1,1)
      le_A(x1,x2) = (2,1,0,1)
      app_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,1,0)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,0,0),(1,1,0,0)) x2 + (0,1,1,1)
      nil_A() = (0,0,0,2)
      min_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x1 + (1,4,3,15)
      if_min_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,0,0,1)) x2
      rm_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(1,0,0,0),(1,0,0,0)) x2 + (0,1,5,5)
      if_rm_A(x1,x2,x3) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,0,1)) x3 + (0,2,1,1)
      minsort_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + x2 + (1,0,7,1)
      if_minsort_A(x1,x2,x3) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,0)) x2 + x3 + (1,0,0,1)
      rand_A(x1) = x1 + (1,1,1,0)

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-minimaldependency 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      eq#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,1,0)) x1 + x2
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(1,1,1,1)) x1 + (0,1,1,1)
      eq_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,0)) x2 + (2,1,1,1)
      |0|_A() = (1,3,2,1)
      true_A() = (1,10,5,2)
      false_A() = (1,3,3,7)
      le_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(1,0,0,0),(1,1,0,0)) x2 + (2,1,6,2)
      app_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x1 + x2 + (0,0,1,2)
      nil_A() = (0,3,0,0)
      add_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (3,1,2,1)
      min_A(x1) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x1 + (3,11,12,12)
      if_min_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,1,0,1)) x1 + ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,1,0,0)) x2
      rm_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + x2 + (0,3,4,1)
      if_rm_A(x1,x2,x3) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + x3 + (0,1,1,0)
      minsort_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + (1,2,1,0)
      if_minsort_A(x1,x2,x3) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,0,0,1)) x2 + ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,1,0)) x3 + (1,0,0,0)
      rand_A(x1) = x1 + (1,0,1,0)

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimaldependency 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      if_min#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,0,1)) x1 + ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2
      false_A() = (1,15,7,4)
      add_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,1,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + (12,1,2,1)
      min#_A(x1) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (12,5,32,5)
      le_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x2 + (11,9,5,1)
      true_A() = (10,8,9,11)
      eq_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (6,1,1,5)
      |0|_A() = (5,1,1,1)
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(1,1,1,0)) x1 + (0,15,0,1)
      app_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,1,0)) x2 + (0,2,0,0)
      nil_A() = (0,1,0,2)
      min_A(x1) = x1 + (12,0,2,4)
      if_min_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,1,0)) x1 + x2
      rm_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,0,0,0)) x2 + (0,3,1,1)
      if_rm_A(x1,x2,x3) = x3 + (0,2,0,0)
      minsort_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + x2 + (1,2,4,1)
      if_minsort_A(x1,x2,x3) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x2 + x3 + (1,1,3,3)
      rand_A(x1) = x1 + (1,0,1,0)

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-minimaldependency 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      le#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,1,1,0),(1,1,0,1)) x2
      s_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(1,1,1,0)) x1 + (0,1,1,1)
      eq_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,0)) x2 + (2,1,1,1)
      |0|_A() = (1,0,1,1)
      true_A() = (1,0,4,4)
      false_A() = (1,3,4,3)
      le_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,0,1,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,0,1,0)) x2 + (7,3,5,2)
      app_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(1,1,1,0),(0,0,0,1)) x2 + (0,5,1,0)
      nil_A() = (0,1,0,1)
      add_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,1,1,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,1,0)) x2 + (3,1,1,4)
      min_A(x1) = x1 + (1,2,6,7)
      if_min_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,0)) x2
      rm_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,0,0,0)) x2 + (0,5,0,2)
      if_rm_A(x1,x2,x3) = ((0,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,0,1,0)) x3 + (0,0,3,0)
      minsort_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + (1,2,0,1)
      if_minsort_A(x1,x2,x3) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x3 + (1,2,0,0)
      rand_A(x1) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,0,1,0)

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimaldependency 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = x1
      add_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,0,1,0)) x2 + (2,1,1,0)
      eq_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,0,0,0)) x1 + x2 + (1,1,1,1)
      |0|_A() = (2,1,1,1)
      true_A() = (3,5,3,4)
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(1,1,1,0)) x1 + (0,5,3,1)
      false_A() = (1,5,3,2)
      le_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(0,1,0,0)) x1 + x2 + (5,5,1,1)
      app_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,0)) x2 + (0,1,1,2)
      nil_A() = (0,1,0,0)
      min_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + (1,7,0,4)
      if_min_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,1,0,0),(1,1,0,0)) x2
      rm_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x2 + (0,3,1,1)
      if_rm_A(x1,x2,x3) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x2 + ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,1,1,0)) x3 + (0,1,5,0)
      minsort_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(1,0,0,0)) x1 + x2 + (1,2,2,1)
      if_minsort_A(x1,x2,x3) = ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x2 + x3 + (1,1,0,0)
      rand_A(x1) = x1 + (1,0,1,0)

The next rules are strictly ordered:

  p1

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