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, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ifinter#_A(x1,x2,x3,x4) = max{0, x1 - 2, x2 - 103, x3 - 19, x4 - 29}
            false_A = 1
            inter#_A(x1,x2) = max{0, x1 - 29, x2 - 29}
            cons_A(x1,x2) = max{27, x1 - 74, x2 + 10}
            mem_A(x1,x2) = max{1, x1 - 121, x2 - 29}
            true_A = 0
            app_A(x1,x2) = max{30, x1 + 16, x2 + 17}
            if_A(x1,x2,x3) = max{x1, x2, x3}
            eq_A(x1,x2) = max{1, x1 - 119, x2 - 104}
            |0|_A = 101
            s_A(x1) = max{103, x1 + 1}
            nil_A = 17
            ifmem_A(x1,x2,x3) = max{1, x1 - 2, x2 - 121, x3 - 19}
            inter_A(x1,x2) = max{20, x1 - 50, x2 - 62}
            ifinter#!_A(x1,x2,x3,x4) = max{x1 + 4, x3 - 11, x4 - 20}
            false!_A = 3
            inter#!_A(x1,x2) = max{4, x1 - 11, x2 - 20}
            cons!_A(x1,x2) = max{9, x1 - 91, x2 - 120}
            mem!_A(x1,x2) = max{10, x1 - 102, x2 - 28}
            true!_A = 1
            app!_A(x1,x2) = max{10, x1 - 19}
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            eq!_A(x1,x2) = max{10, x1 - 102, x2 - 2}
            |0|!_A = 4
            s!_A(x1) = 4
            nil!_A = 0
            ifmem!_A(x1,x2,x3) = max{10, x1 - 104, x2 - 102, x3 - 28}
            inter!_A(x1,x2) = max{2, x1 - 18, x2 - 98}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ifinter#_A(x1,x2,x3,x4) = max{0, x1 - 1, x2 - 23, x3 - 4, x4 - 21}
            false_A = 0
            inter#_A(x1,x2) = max{0, x1 - 21, x2 - 21}
            cons_A(x1,x2) = max{19, x1 - 2, x2 + 17}
            mem_A(x1,x2) = max{0, x1 - 22, x2 - 22}
            true_A = 0
            app_A(x1,x2) = max{4, x1 + 1, x2 + 2}
            if_A(x1,x2,x3) = max{x1, x2, x3}
            eq_A(x1,x2) = max{0, x1 - 34, x2 - 22}
            |0|_A = 0
            s_A(x1) = max{33, x1 + 21}
            nil_A = 21
            ifmem_A(x1,x2,x3) = max{0, x1 - 2, x2 - 22, x3 - 5}
            inter_A(x1,x2) = max{29, x1 - 24, x2 - 43}
            ifinter#!_A(x1,x2,x3,x4) = max{4, x1 - 11, x3 - 31, x4 - 12}
            false!_A = 5
            inter#!_A(x1,x2) = max{4, x1 - 31, x2 - 12}
            cons!_A(x1,x2) = max{5, x1 - 34, x2 - 65}
            mem!_A(x1,x2) = max{8, x1 - 1, x2 - 22}
            true!_A = 3
            app!_A(x1,x2) = max{1, x1 - 58, x2 - 30}
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            eq!_A(x1,x2) = max{8, x1 - 34, x2 - 24}
            |0|!_A = 6
            s!_A(x1) = max{6, x1 - 24}
            nil!_A = 0
            ifmem!_A(x1,x2,x3) = max{8, x1 - 2, x2 - 1, x3 - 22}
            inter!_A(x1,x2) = max{2, x1 - 57, x2 - 108}
        
  
    partial status:
  
      pi(ifinter#) = []
      pi(false) = []
      pi(inter#) = []
      pi(cons) = [1, 2]
      pi(mem) = [1]
      pi(true) = []
      pi(app) = [1]
      pi(if) = [2, 3]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(nil) = []
      pi(ifmem) = [2]
      pi(inter) = [1, 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, r10, r11, r12, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ifinter#_A(x1,x2,x3,x4) = max{0, x1 - 12, x2 - 37, x3 - 2, x4 - 2}
            false_A = 0
            inter#_A(x1,x2) = max{0, x1 - 19, x2 - 2}
            app_A(x1,x2) = max{14, x1 + 12, x2 + 13}
            cons_A(x1,x2) = max{x1 - 18, x2 + 17}
            mem_A(x1,x2) = max{11, x1 - 25, x2 - 40}
            true_A = 2
            if_A(x1,x2,x3) = max{x1, x2, x3}
            eq_A(x1,x2) = max{2, x1 - 1, x2 - 34}
            |0|_A = 0
            s_A(x1) = x1
            nil_A = 2
            ifmem_A(x1,x2,x3) = max{11, x1 - 24, x2 - 25, x3 - 23}
            ifinter#!_A(x1,x2,x3,x4) = max{8, x3 - 1}
            false!_A = 4
            inter#!_A(x1,x2) = max{5, x1 - 8}
            app!_A(x1,x2) = max{3, x1 - 20, x2 - 107}
            cons!_A(x1,x2) = max{1, x1 - 122, x2 - 122}
            mem!_A(x1,x2) = max{10, x1 - 25, x2 - 3}
            true!_A = 2
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            eq!_A(x1,x2) = max{6, x1 - 26, x2 - 26}
            |0|!_A = 5
            s!_A(x1) = max{5, x1 - 26}
            nil!_A = 2
            ifmem!_A(x1,x2,x3) = max{14, x1 - 24, x2 - 25, x3 - 3}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ifinter#_A(x1,x2,x3,x4) = max{0, x1 - 15, x2 - 17, x3 - 3, x4 - 16}
            false_A = 0
            inter#_A(x1,x2) = max{0, x1 - 6, x2 - 16}
            app_A(x1,x2) = max{x1 + 1, x2 + 4}
            cons_A(x1,x2) = max{5, x1 - 11, x2 + 3}
            mem_A(x1,x2) = max{0, x1 - 2, x2 - 8}
            true_A = 0
            if_A(x1,x2,x3) = max{x1, x2, x3}
            eq_A(x1,x2) = max{42, x1 - 21, x2 - 21}
            |0|_A = 1
            s_A(x1) = max{20, x1 + 1}
            nil_A = 7
            ifmem_A(x1,x2,x3) = max{0, x1 - 43, x2 - 2, x3 - 5}
            ifinter#!_A(x1,x2,x3,x4) = max{1, x1 - 42, x2 - 1}
            false!_A = 18
            inter#!_A(x1,x2) = max{17, x1 + 13, x2 - 9}
            app!_A(x1,x2) = max{8, x1 - 34, x2 - 10}
            cons!_A(x1,x2) = max{5, x1 - 45, x2 - 41}
            mem!_A(x1,x2) = max{19, x1 + 3, x2 - 8}
            true!_A = 18
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            eq!_A(x1,x2) = max{4, x1 - 2, x2 - 21}
            |0|!_A = 19
            s!_A(x1) = max{1, x1 - 22}
            nil!_A = 0
            ifmem!_A(x1,x2,x3) = max{19, x1 - 23, x2 + 3, x3 - 6}
        
  
    partial status:
  
      pi(ifinter#) = []
      pi(false) = []
      pi(inter#) = []
      pi(app) = [1]
      pi(cons) = []
      pi(mem) = [1]
      pi(true) = []
      pi(if) = [2, 3]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(nil) = []
      pi(ifmem) = [2]

The following rules are removed:

  p1, p2, p3, p6, p7



-- SCC decomposition.

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

p1: inter#(l1,app(l2,l3)) -> inter#(l1,l2)
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            inter#_A(x1,x2) = 0
            app_A(x1,x2) = max{3, x1 - 29, x2 - 28}
            if_A(x1,x2,x3) = max{x1, x2, x3}
            true_A = 5
            false_A = 3
            inter#!_A(x1,x2) = max{1, x2 - 31}
            app!_A(x1,x2) = max{6, x1 - 6, x2 - 30}
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            true!_A = 4
            false!_A = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            inter#_A(x1,x2) = 0
            app_A(x1,x2) = max{13, x1 - 17, x2 - 14}
            if_A(x1,x2,x3) = max{x1, x2, x3}
            true_A = 2
            false_A = 2
            inter#!_A(x1,x2) = max{8, x1 - 1, x2 - 17}
            app!_A(x1,x2) = max{7, x1 - 17, x2 - 28}
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            true!_A = 5
            false!_A = 3
        
  
    partial status:
  
      pi(inter#) = [2]
      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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ifmem#_A(x1,x2,x3) = max{0, x1 - 3, x2 - 2, x3 - 5}
            false_A = 2
            mem#_A(x1,x2) = max{0, x1 - 2, x2 - 7}
            cons_A(x1,x2) = max{x1 + 5, x2 + 2}
            eq_A(x1,x2) = max{2, x1 - 11, x2 - 11}
            if_A(x1,x2,x3) = max{x1, x2, x3}
            true_A = 2
            |0|_A = 1
            s_A(x1) = max{10, x1 + 1}
            ifmem#!_A(x1,x2,x3) = max{8, x1 - 56, x2 - 1, x3 - 4}
            false!_A = 5
            mem#!_A(x1,x2) = max{8, x1 - 1, x2 - 6}
            cons!_A(x1,x2) = max{2, x1 - 53, x2 - 38}
            eq!_A(x1,x2) = max{7, x1 - 11, x2 - 2}
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            true!_A = 1
            |0|!_A = 6
            s!_A(x1) = max{6, x1 - 3}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ifmem#_A(x1,x2,x3) = max{0, x1 - 3, x2 - 2, x3 - 15}
            false_A = 2
            mem#_A(x1,x2) = max{0, x1 - 2, x2 - 19}
            cons_A(x1,x2) = max{x1 + 2, x2 + 4}
            eq_A(x1,x2) = max{2, x1 - 12, x2 - 14}
            if_A(x1,x2,x3) = max{x1, x2, x3}
            true_A = 2
            |0|_A = 11
            s_A(x1) = x1 + 1
            ifmem#!_A(x1,x2,x3) = max{4, x1 - 4, x2 - 1, x3 - 5}
            false!_A = 1
            mem#!_A(x1,x2) = max{4, x1 - 1, x2 - 5}
            cons!_A(x1,x2) = max{5, x1 - 58, x2 - 60}
            eq!_A(x1,x2) = max{3, x1 - 2, x2 - 59}
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            true!_A = 8
            |0|!_A = 9
            s!_A(x1) = max{2, x1 - 3}
        
  
    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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            eq#_A(x1,x2) = max{x1, x2}
            s_A(x1) = x1
            if_A(x1,x2,x3) = max{x1, x2, x3}
            true_A = 2
            false_A = 2
            eq#!_A(x1,x2) = max{2, x1 - 1, x2 - 3}
            s!_A(x1) = max{1, x1 - 2}
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            true!_A = 3
            false!_A = 0
        
        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 + 1
            if_A(x1,x2,x3) = max{x1, x2, x3}
            true_A = 1
            false_A = 2
            eq#!_A(x1,x2) = max{3, x1 - 33, x2 - 33}
            s!_A(x1) = max{2, x1 - 33}
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            true!_A = 4
            false!_A = 1
        
  
    partial status:
  
      pi(eq#) = [1]
      pi(s) = [1]
      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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 10, x2 - 13}
            cons_A(x1,x2) = max{4, x1 - 25, x2}
            app_A(x1,x2) = max{9, x1 + 3, x2}
            if_A(x1,x2,x3) = max{x1, x2, x3}
            true_A = 1
            false_A = 13
            nil_A = 21
            app#!_A(x1,x2) = max{2, x1 - 10, x2 - 20}
            cons!_A(x1,x2) = max{1, x1 - 73, x2 - 22}
            app!_A(x1,x2) = max{4, x1 - 18, x2 - 21}
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            true!_A = 2
            false!_A = 4
            nil!_A = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 14, x2 - 17}
            cons_A(x1,x2) = max{4, x1 - 54, x2}
            app_A(x1,x2) = max{13, x1 + 3, x2}
            if_A(x1,x2,x3) = max{x1, x2, x3}
            true_A = 1
            false_A = 2
            nil_A = 7
            app#!_A(x1,x2) = max{13, x1 - 14, x2 - 17}
            cons!_A(x1,x2) = max{12, x1 - 84, x2 - 27}
            app!_A(x1,x2) = max{20, x1 - 15, x2 - 26}
            if!_A(x1,x2,x3) = max{x1, x2, x3}
            true!_A = 14
            false!_A = 2
            nil!_A = 5
        
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [2]
      pi(app) = [1, 2]
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(nil) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.