YES

We show the termination of the TRS 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))

-- SCC decomposition.

Consider the 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))

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

Consider the 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))

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:

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

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      if_rm#_A(x1,x2,x3) = ((1,0),(1,0)) x3 + (0,1)
      false_A() = (1,3)
      add_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,1)
      rm#_A(x1,x2) = ((1,0),(1,0)) x2 + (1,0)
      eq_A(x1,x2) = (2,2)
      true_A() = (1,1)
      |0|_A() = (1,1)
      s_A(x1) = ((0,0),(1,0)) x1 + (1,1)

The next rules are strictly ordered:

  p1, p2, p3

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

-- Reduction pair.

Consider the 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))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  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) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21

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: 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))

The set of usable rules consists of

  r5, r6, r7

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      if_min#_A(x1,x2) = x2 + (0,1)
      false_A() = (1,4)
      add_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,0)
      min#_A(x1) = ((1,0),(1,0)) x1 + (1,0)
      le_A(x1,x2) = x1 + x2 + (1,1)
      true_A() = (1,1)
      |0|_A() = (1,1)
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1, p2, p3

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

-- Reduction pair.

Consider the 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))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  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) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21

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: 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))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0),(1,0)) x1
      add_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)

The next rules are strictly ordered:

  p1

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