YES

We show the termination of the TRS R:

  if(true(),x,y) -> x
  if(false(),x,y) -> y
  eq(|0|(),|0|()) -> true()
  eq(|0|(),s(x)) -> false()
  eq(s(x),|0|()) -> false()
  eq(s(x),s(y)) -> eq(x,y)
  app(nil(),l) -> l
  app(cons(x,l1),l2) -> cons(x,app(l1,l2))
  app(app(l1,l2),l3) -> app(l1,app(l2,l3))
  mem(x,nil()) -> false()
  mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
  ifmem(true(),x,l) -> true()
  ifmem(false(),x,l) -> mem(x,l)
  inter(x,nil()) -> nil()
  inter(nil(),x) -> nil()
  inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
  inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
  inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
  inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
  ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
  ifinter(false(),x,l1,l2) -> inter(l1,l2)

-- SCC decomposition.

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

p1: eq#(s(x),s(y)) -> eq#(x,y)
p2: app#(cons(x,l1),l2) -> app#(l1,l2)
p3: app#(app(l1,l2),l3) -> app#(l1,app(l2,l3))
p4: app#(app(l1,l2),l3) -> app#(l2,l3)
p5: mem#(x,cons(y,l)) -> ifmem#(eq(x,y),x,l)
p6: mem#(x,cons(y,l)) -> eq#(x,y)
p7: ifmem#(false(),x,l) -> mem#(x,l)
p8: inter#(app(l1,l2),l3) -> app#(inter(l1,l3),inter(l2,l3))
p9: inter#(app(l1,l2),l3) -> inter#(l1,l3)
p10: inter#(app(l1,l2),l3) -> inter#(l2,l3)
p11: inter#(l1,app(l2,l3)) -> app#(inter(l1,l2),inter(l1,l3))
p12: inter#(l1,app(l2,l3)) -> inter#(l1,l2)
p13: inter#(l1,app(l2,l3)) -> inter#(l1,l3)
p14: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p15: inter#(cons(x,l1),l2) -> mem#(x,l2)
p16: inter#(l1,cons(x,l2)) -> ifinter#(mem(x,l1),x,l2,l1)
p17: inter#(l1,cons(x,l2)) -> mem#(x,l1)
p18: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p19: ifinter#(false(),x,l1,l2) -> inter#(l1,l2)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The estimated dependency graph contains the following SCCs:

  {p9, p10, p12, p13, p14, p16, p18, p19}
  {p5, p7}
  {p1}
  {p2, p3, p4}


-- Reduction pair.

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

p1: ifinter#(false(),x,l1,l2) -> inter#(l1,l2)
p2: inter#(l1,cons(x,l2)) -> ifinter#(mem(x,l1),x,l2,l1)
p3: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p4: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p5: inter#(l1,app(l2,l3)) -> inter#(l1,l3)
p6: inter#(l1,app(l2,l3)) -> inter#(l1,l2)
p7: inter#(app(l1,l2),l3) -> inter#(l2,l3)
p8: inter#(app(l1,l2),l3) -> inter#(l1,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  r3, r4, r5, r6, r10, r11, r12, r13

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    ifinter#_A(x1,x2,x3,x4) = max{x1 + 2, x3 + 4, x4 + 3}
    false_A = 0
    inter#_A(x1,x2) = max{x1 + 4, x2 + 3}
    cons_A(x1,x2) = max{x1, x2 + 2}
    mem_A(x1,x2) = 0
    true_A = 0
    app_A(x1,x2) = max{x1, x2}
    eq_A(x1,x2) = 2
    |0|_A = 0
    s_A(x1) = max{0, x1 - 1}
    ifmem_A(x1,x2,x3) = 0
    nil_A = 0

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: ifinter#(false(),x,l1,l2) -> inter#(l1,l2)
p2: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p3: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p4: inter#(l1,app(l2,l3)) -> inter#(l1,l3)
p5: inter#(l1,app(l2,l3)) -> inter#(l1,l2)
p6: inter#(app(l1,l2),l3) -> inter#(l2,l3)
p7: inter#(app(l1,l2),l3) -> inter#(l1,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

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

p1: ifinter#(false(),x,l1,l2) -> inter#(l1,l2)
p2: inter#(app(l1,l2),l3) -> inter#(l1,l3)
p3: inter#(app(l1,l2),l3) -> inter#(l2,l3)
p4: inter#(l1,app(l2,l3)) -> inter#(l1,l2)
p5: inter#(l1,app(l2,l3)) -> inter#(l1,l3)
p6: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p7: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  r3, r4, r5, r6, r10, r11, r12, r13

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    ifinter#_A(x1,x2,x3,x4) = max{x1 + 2, x4 + 5}
    false_A = 3
    inter#_A(x1,x2) = x2 + 5
    app_A(x1,x2) = max{x1 + 1, x2}
    cons_A(x1,x2) = max{0, x1 - 5, x2 - 3}
    mem_A(x1,x2) = 3
    true_A = 0
    eq_A(x1,x2) = max{5, x1, x2 - 2}
    |0|_A = 5
    s_A(x1) = x1 + 5
    ifmem_A(x1,x2,x3) = 3
    nil_A = 0

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: ifinter#(false(),x,l1,l2) -> inter#(l1,l2)
p2: inter#(app(l1,l2),l3) -> inter#(l1,l3)
p3: inter#(app(l1,l2),l3) -> inter#(l2,l3)
p4: inter#(l1,app(l2,l3)) -> inter#(l1,l3)
p5: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p6: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- Reduction pair.

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

p1: ifinter#(false(),x,l1,l2) -> inter#(l1,l2)
p2: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p3: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p4: inter#(l1,app(l2,l3)) -> inter#(l1,l3)
p5: inter#(app(l1,l2),l3) -> inter#(l2,l3)
p6: inter#(app(l1,l2),l3) -> inter#(l1,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  r3, r4, r5, r6, r10, r11, r12, r13

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    ifinter#_A(x1,x2,x3,x4) = max{x1 + 6, x2 + 6, x3 + 6}
    false_A = 1
    inter#_A(x1,x2) = max{6, x1 + 5}
    cons_A(x1,x2) = max{x1 + 2, x2 + 2}
    mem_A(x1,x2) = 1
    true_A = 0
    app_A(x1,x2) = max{1, x1, x2}
    eq_A(x1,x2) = 3
    |0|_A = 0
    s_A(x1) = max{0, x1 - 1}
    ifmem_A(x1,x2,x3) = 1
    nil_A = 0

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: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p2: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p3: inter#(l1,app(l2,l3)) -> inter#(l1,l3)
p4: inter#(app(l1,l2),l3) -> inter#(l2,l3)
p5: inter#(app(l1,l2),l3) -> inter#(l1,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

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: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p2: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p3: inter#(app(l1,l2),l3) -> inter#(l1,l3)
p4: inter#(app(l1,l2),l3) -> inter#(l2,l3)
p5: inter#(l1,app(l2,l3)) -> inter#(l1,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  r3, r4, r5, r6, r10, r11, r12, r13

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    inter#_A(x1,x2) = x2 + 1
    cons_A(x1,x2) = max{0, x1 - 4, x2 - 3}
    ifinter#_A(x1,x2,x3,x4) = max{x1 + 1, x4 + 1}
    mem_A(x1,x2) = 0
    true_A = 0
    app_A(x1,x2) = max{x1 - 1, x2 + 1}
    eq_A(x1,x2) = 3
    |0|_A = 0
    s_A(x1) = max{0, x1 - 1}
    false_A = 0
    ifmem_A(x1,x2,x3) = 0
    nil_A = 0

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p2: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p3: inter#(app(l1,l2),l3) -> inter#(l1,l3)
p4: inter#(app(l1,l2),l3) -> inter#(l2,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

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: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p2: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p3: inter#(app(l1,l2),l3) -> inter#(l2,l3)
p4: inter#(app(l1,l2),l3) -> inter#(l1,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  r3, r4, r5, r6, r10, r11, r12, r13

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    inter#_A(x1,x2) = x1 + 2
    cons_A(x1,x2) = max{x1, x2}
    ifinter#_A(x1,x2,x3,x4) = max{x1 + 2, x2 + 1, x3 + 2}
    mem_A(x1,x2) = 0
    true_A = 0
    app_A(x1,x2) = max{x1 + 1, x2}
    eq_A(x1,x2) = 2
    |0|_A = 0
    s_A(x1) = max{0, x1 - 1}
    false_A = 0
    ifmem_A(x1,x2,x3) = 0
    nil_A = 2

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p2: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p3: inter#(app(l1,l2),l3) -> inter#(l2,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

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: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,l1,l2)
p2: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p3: inter#(app(l1,l2),l3) -> inter#(l2,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  r3, r4, r5, r6, r10, r11, r12, r13

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    inter#_A(x1,x2) = x1 + 6
    cons_A(x1,x2) = max{x1 - 2, x2 + 1}
    ifinter#_A(x1,x2,x3,x4) = max{x2 + 1, x3 + 6}
    mem_A(x1,x2) = max{x1 + 3, x2 + 3}
    true_A = 3
    app_A(x1,x2) = max{x1 - 5, x2}
    eq_A(x1,x2) = max{x1 + 4, x2}
    |0|_A = 3
    s_A(x1) = x1 + 3
    false_A = 3
    ifmem_A(x1,x2,x3) = max{x1 - 1, x2 + 3, x3 + 3}
    nil_A = 0

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: ifinter#(true(),x,l1,l2) -> inter#(l1,l2)
p2: inter#(app(l1,l2),l3) -> inter#(l2,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: inter#(app(l1,l2),l3) -> inter#(l2,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    inter#_A(x1,x2) = x1
    app_A(x1,x2) = max{x1 + 1, x2 + 1}

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: ifmem#(false(),x,l) -> mem#(x,l)
p2: mem#(x,cons(y,l)) -> ifmem#(eq(x,y),x,l)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  r3, r4, r5, r6

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    ifmem#_A(x1,x2,x3) = max{x1 - 2, x3 + 1}
    false_A = 0
    mem#_A(x1,x2) = max{1, x2}
    cons_A(x1,x2) = max{x1 + 1, x2 + 2}
    eq_A(x1,x2) = max{0, x2 - 1}
    |0|_A = 0
    true_A = 0
    s_A(x1) = x1

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: ifmem#(false(),x,l) -> mem#(x,l)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    eq#_A(x1,x2) = max{x1 + 1, x2 + 1}
    s_A(x1) = x1 + 1

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: app#(cons(x,l1),l2) -> app#(l1,l2)
p2: app#(app(l1,l2),l3) -> app#(l2,l3)
p3: app#(app(l1,l2),l3) -> app#(l1,app(l2,l3))

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  r7, r8, r9

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    app#_A(x1,x2) = x1 + 4
    cons_A(x1,x2) = max{2, x1, x2}
    app_A(x1,x2) = max{x1 + 2, x2}
    nil_A = 0

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: app#(cons(x,l1),l2) -> app#(l1,l2)
p2: app#(app(l1,l2),l3) -> app#(l2,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: app#(cons(x,l1),l2) -> app#(l1,l2)
p2: app#(app(l1,l2),l3) -> app#(l2,l3)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    app#_A(x1,x2) = max{0, x1 - 1}
    cons_A(x1,x2) = x2 + 1
    app_A(x1,x2) = max{x1 + 2, x2 + 2}

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: app#(cons(x,l1),l2) -> app#(l1,l2)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: app#(cons(x,l1),l2) -> app#(l1,l2)

and R consists of:

r1: if(true(),x,y) -> x
r2: if(false(),x,y) -> y
r3: eq(|0|(),|0|()) -> true()
r4: eq(|0|(),s(x)) -> false()
r5: eq(s(x),|0|()) -> false()
r6: eq(s(x),s(y)) -> eq(x,y)
r7: app(nil(),l) -> l
r8: app(cons(x,l1),l2) -> cons(x,app(l1,l2))
r9: app(app(l1,l2),l3) -> app(l1,app(l2,l3))
r10: mem(x,nil()) -> false()
r11: mem(x,cons(y,l)) -> ifmem(eq(x,y),x,l)
r12: ifmem(true(),x,l) -> true()
r13: ifmem(false(),x,l) -> mem(x,l)
r14: inter(x,nil()) -> nil()
r15: inter(nil(),x) -> nil()
r16: inter(app(l1,l2),l3) -> app(inter(l1,l3),inter(l2,l3))
r17: inter(l1,app(l2,l3)) -> app(inter(l1,l2),inter(l1,l3))
r18: inter(cons(x,l1),l2) -> ifinter(mem(x,l2),x,l1,l2)
r19: inter(l1,cons(x,l2)) -> ifinter(mem(x,l1),x,l2,l1)
r20: ifinter(true(),x,l1,l2) -> cons(x,inter(l1,l2))
r21: ifinter(false(),x,l1,l2) -> inter(l1,l2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    app#_A(x1,x2) = x1
    cons_A(x1,x2) = max{x1 + 1, x2 + 1}

The next rules are strictly ordered:

  p1

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