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

  r1, r2, r3, r4, r5, r6, r10, r11, r12, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ifinter#_A(x1,x2,x3,x4) = 1
        false_A = 4
        inter#_A(x1,x2) = 1
        cons_A(x1,x2) = max{5, x1 - 6, x2 - 6}
        mem_A(x1,x2) = x1 + 5
        true_A = 4
        app_A(x1,x2) = max{0, x1 - 6, x2 - 6}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        eq_A(x1,x2) = max{6, x1 - 6}
        |0|_A = 5
        s_A(x1) = x1 + 13
        nil_A = 5
        ifmem_A(x1,x2,x3) = max{x1 - 7, x2 + 5}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          ifinter#_A(x1,x2,x3,x4) = max{0, x1 - 97, x2 - 102, x3 - 1, x4 - 98}
          false_A = 0
          inter#_A(x1,x2) = max{0, x1 - 98, x2 - 98}
          cons_A(x1,x2) = max{x1 - 4, x2 + 97}
          mem_A(x1,x2) = max{0, x1 - 5, x2 - 44}
          true_A = 53
          app_A(x1,x2) = max{x1 + 2, x2 + 39}
          if_A(x1,x2,x3) = max{x1, x2, x3}
          eq_A(x1,x2) = max{53, x1 - 2, x2 - 2}
          |0|_A = 0
          s_A(x1) = x1 + 1
          nil_A = 42
          ifmem_A(x1,x2,x3) = max{x1 - 54, x2 - 5, x3 + 53}
          ifinter#!_A(x1,x2,x3,x4) = max{123, x1 - 54, x2 - 1, x3 + 83, x4 + 82}
          false!_A = 1
          inter#!_A(x1,x2) = max{122, x1 + 83, x2 + 82}
          cons!_A(x1,x2) = max{76, x1 - 27, x2 - 51}
          mem!_A(x1,x2) = x1 + 80
          true!_A = 75
          app!_A(x1,x2) = max{1, x1 - 22, x2 - 44}
          if!_A(x1,x2,x3) = max{x1, x2, x3}
          eq!_A(x1,x2) = max{81, x1 - 2, x2 - 2}
          |0|!_A = 76
          s!_A(x1) = max{2, x1 - 3}
          nil!_A = 77
          ifmem!_A(x1,x2,x3) = x2 + 80
  
    partial status:
  
      pi(ifinter#) = []
      pi(false) = []
      pi(inter#) = []
      pi(cons) = []
      pi(mem) = [1]
      pi(true) = []
      pi(app) = []
      pi(if) = [2, 3]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(nil) = []
      pi(ifmem) = [2]

The following rules are removed:

  p2



-- 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

  r1, r2, r3, r4, r5, r6, r7, r10, r11, r12, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ifinter#_A(x1,x2,x3,x4) = x4 + 19
        false_A = 4
        inter#_A(x1,x2) = x2 + 19
        app_A(x1,x2) = max{5, x1 + 3, x2 + 4}
        cons_A(x1,x2) = max{6, x1 - 9, x2 - 9}
        mem_A(x1,x2) = max{8, x1 - 10}
        true_A = 4
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        eq_A(x1,x2) = max{7, x1 - 11}
        |0|_A = 5
        s_A(x1) = x1 + 8
        nil_A = 0
        ifmem_A(x1,x2,x3) = max{8, x1 - 8, x2 - 10}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          ifinter#_A(x1,x2,x3,x4) = max{0, x1 - 2, x2 - 2, x3 - 2, x4 - 2}
          false_A = 1
          inter#_A(x1,x2) = max{0, x1 - 27, x2 - 2}
          app_A(x1,x2) = max{14, x1 + 1, x2 + 2}
          cons_A(x1,x2) = max{x1 + 25, x2 + 25}
          mem_A(x1,x2) = max{1, x1 - 83}
          true_A = 1
          if_A(x1,x2,x3) = max{x1, x2, x3}
          eq_A(x1,x2) = max{1, x1 - 2}
          |0|_A = 1
          s_A(x1) = x1 + 1
          nil_A = 25
          ifmem_A(x1,x2,x3) = max{1, x1 - 81, x2 - 83}
          ifinter#!_A(x1,x2,x3,x4) = max{x1 + 25, x2 + 28, x3 + 28}
          false!_A = 21
          inter#!_A(x1,x2) = max{27, x1 + 4}
          app!_A(x1,x2) = max{26, x1 - 32, x2 - 43}
          cons!_A(x1,x2) = max{25, x1 - 58, x2 - 56}
          mem!_A(x1,x2) = max{24, x1 - 5, x2 - 26}
          true!_A = 1
          if!_A(x1,x2,x3) = max{x1, x2, x3}
          eq!_A(x1,x2) = max{23, x1 - 6, x2 - 6}
          |0|!_A = 22
          s!_A(x1) = max{22, x1 - 6}
          nil!_A = 0
          ifmem!_A(x1,x2,x3) = max{24, x1 - 3, x2 - 5, x3 - 26}
  
    partial status:
  
      pi(ifinter#) = []
      pi(false) = []
      pi(inter#) = [2]
      pi(app) = []
      pi(cons) = []
      pi(mem) = []
      pi(true) = []
      pi(if) = [2, 3]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(nil) = []
      pi(ifmem) = []

The following rules are removed:

  p1, p4, p5, p6, p7



-- SCC decomposition.

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

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

  {p1, p2}


-- Reduction pair.

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

p1: inter#(app(l1,l2),l3) -> inter#(l1,l3)
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 set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        inter#_A(x1,x2) = max{x1 + 2, x2 + 20}
        app_A(x1,x2) = max{17, x1 + 1, x2 + 1}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 1
        false_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          inter#_A(x1,x2) = 0
          app_A(x1,x2) = max{10, x1 - 26, x2 - 26}
          if_A(x1,x2,x3) = max{x1, x2, x3}
          true_A = 1
          false_A = 2
          inter#!_A(x1,x2) = max{0, x2 - 1}
          app!_A(x1,x2) = max{6, x1 - 12}
          if!_A(x1,x2,x3) = max{x1, x2, x3}
          true!_A = 4
          false!_A = 1
  
    partial status:
  
      pi(inter#) = [1]
      pi(app) = [1, 2]
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []

The following rules are removed:

  p1, p2

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

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ifmem#_A(x1,x2,x3) = max{16, x1 - 4, x2 - 4, x3 + 1}
        false_A = 1
        mem#_A(x1,x2) = max{16, x1 - 4, x2 + 1}
        cons_A(x1,x2) = max{14, x1 + 12, x2 + 12}
        eq_A(x1,x2) = max{3, x1 - 5, x2 - 5}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 1
        |0|_A = 2
        s_A(x1) = x1 + 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          ifmem#_A(x1,x2,x3) = max{0, x1 - 10, x2 - 9, x3 - 5}
          false_A = 2
          mem#_A(x1,x2) = max{0, x1 - 9, x2 - 6}
          cons_A(x1,x2) = max{5, x1 - 6, x2 + 1}
          eq_A(x1,x2) = max{2, x1 - 2, x2 - 2}
          if_A(x1,x2,x3) = max{x1, x2, x3}
          true_A = 2
          |0|_A = 0
          s_A(x1) = x1 + 1
          ifmem#!_A(x1,x2,x3) = max{x1 - 15, x2 + 8, x3 - 6}
          false!_A = 1
          mem#!_A(x1,x2) = max{x1 + 8, x2 - 6}
          cons!_A(x1,x2) = max{6, x1 - 5, x2 - 7}
          eq!_A(x1,x2) = max{2, x1 - 35, x2 - 35}
          if!_A(x1,x2,x3) = max{x1, x2, x3}
          true!_A = 1
          |0|!_A = 2
          s!_A(x1) = max{1, x1 - 36}
  
    partial status:
  
      pi(ifmem#) = [3]
      pi(false) = []
      pi(mem#) = [2]
      pi(cons) = [2]
      pi(eq) = []
      pi(if) = [2, 3]
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p2



-- 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

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        eq#_A(x1,x2) = 1
        s_A(x1) = max{6, x1 - 21}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 2
        false_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          eq#_A(x1,x2) = max{x1, x2}
          s_A(x1) = x1 + 8
          if_A(x1,x2,x3) = max{x1, x2, x3}
          true_A = 1
          false_A = 2
          eq#!_A(x1,x2) = max{3, x1 - 1, x2 - 9}
          s!_A(x1) = max{2, x1 - 60}
          if!_A(x1,x2,x3) = max{x1, x2, x3}
          true!_A = 1
          false!_A = 1
  
    partial status:
  
      pi(eq#) = []
      pi(s) = []
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []

The following rules are removed:

  p1

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

  r1, r2, r7, r8, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = 1
        cons_A(x1,x2) = max{18, x1 - 28, x2 - 14}
        app_A(x1,x2) = max{x1 + 17, x2}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 2
        false_A = 1
        nil_A = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 7, x2 - 8}
          cons_A(x1,x2) = max{4, x1 - 2, x2}
          app_A(x1,x2) = max{6, x1 + 1, x2}
          if_A(x1,x2,x3) = max{x1, x2, x3}
          true_A = 1
          false_A = 2
          nil_A = 9
          app#!_A(x1,x2) = max{11, x1 + 8}
          cons!_A(x1,x2) = max{9, x1 - 33, x2 - 14}
          app!_A(x1,x2) = max{10, x1 - 7, x2 - 14}
          if!_A(x1,x2,x3) = max{x1, x2, x3}
          true!_A = 3
          false!_A = 5
          nil!_A = 6
  
    partial status:
  
      pi(app#) = []
      pi(cons) = []
      pi(app) = [1, 2]
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(nil) = []

The following rules are removed:

  p3



-- 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

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{x1 + 8, x2 + 12}
        cons_A(x1,x2) = max{3, x1 - 13, x2 + 1}
        app_A(x1,x2) = max{10, x1 - 15, x2 + 2}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 1
        false_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = 0
          cons_A(x1,x2) = max{28, x1 - 34, x2 - 34}
          app_A(x1,x2) = max{15, x1 - 48, x2 - 34}
          if_A(x1,x2,x3) = max{x1, x2, x3}
          true_A = 1
          false_A = 2
          app#!_A(x1,x2) = max{0, x2 - 1}
          cons!_A(x1,x2) = max{2, x1 - 108, x2 - 124}
          app!_A(x1,x2) = max{7, x1 - 94, x2 - 110}
          if!_A(x1,x2,x3) = max{x1, x2, x3}
          true!_A = 5
          false!_A = 1
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [2]
      pi(app) = []
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.