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

  r7, r14, r15

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) = max{15, x3 - 4, x4 - 9}
        false_A = 13
        inter#_A(x1,x2) = max{6, x1 - 4, x2 - 9}
        cons_A(x1,x2) = max{24, x2 + 5}
        mem_A(x1,x2) = 14
        true_A = 4
        app_A(x1,x2) = max{8, x1 + 5, x2 + 5}
        nil_A = 6
        inter_A(x1,x2) = 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            ifinter#_A(x1,x2,x3,x4) = x3 + x4
            false_A() = 14
            inter#_A(x1,x2) = x1 + x2
            cons_A(x1,x2) = x1 + x2 + 2
            mem_A(x1,x2) = 1
            true_A() = 14
            app_A(x1,x2) = x1 + x2 + 1
            nil_A() = 9
            inter_A(x1,x2) = x1 + x2 + 1
            ifinter#!_A(x1,x2,x3,x4) = x3 + x4 + 10
            false!_A() = 8
            inter#!_A(x1,x2) = x1 + x2 + 9
            cons!_A(x1,x2) = 4
            mem!_A(x1,x2) = x2
            true!_A() = 5
            app!_A(x1,x2) = x1 + 5
            nil!_A() = 6
            inter!_A(x1,x2) = 7
  
    partial status:
  
      pi(ifinter#) = []
      pi(false) = []
      pi(inter#) = []
      pi(cons) = []
      pi(mem) = []
      pi(true) = []
      pi(app) = []
      pi(nil) = []
      pi(inter) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6, p7, p8

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, r7, r14, r15

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{17, x3 + 3}
        false_A = 1
        mem#_A(x1,x2) = x2 + 2
        cons_A(x1,x2) = max{16, x1 + 8, x2 + 8}
        eq_A(x1,x2) = 3
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 1
        |0|_A = 3
        s_A(x1) = max{4, x1 + 2}
        app_A(x1,x2) = max{17, x1 + 1, x2}
        nil_A = 5
        inter_A(x1,x2) = max{x1 + 4, x2 + 6}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            ifmem#_A(x1,x2,x3) = x1 + x3
            false_A() = 2
            mem#_A(x1,x2) = x2 + 2
            cons_A(x1,x2) = x1 + x2 + 1
            eq_A(x1,x2) = x2 + 1
            if_A(x1,x2,x3) = x2 + x3
            true_A() = 1
            |0|_A() = 1
            s_A(x1) = x1 + 1
            app_A(x1,x2) = x1 + x2 + 1
            nil_A() = 1
            inter_A(x1,x2) = x1 + x2 + 2
            ifmem#!_A(x1,x2,x3) = x1 + x2 + x3 + 3
            false!_A() = 1
            mem#!_A(x1,x2) = x2
            cons!_A(x1,x2) = x1 + x2 + 3
            eq!_A(x1,x2) = 2
            if!_A(x1,x2,x3) = x1 + x2 + x3
            true!_A() = 0
            |0|!_A() = 2
            s!_A(x1) = 3
            app!_A(x1,x2) = x1 + x2 + 2
            nil!_A() = 2
            inter!_A(x1,x2) = x1 + x2 + 4
  
    partial status:
  
      pi(ifmem#) = []
      pi(false) = []
      pi(mem#) = [2]
      pi(cons) = [1]
      pi(eq) = []
      pi(if) = [2, 3]
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(app) = [1, 2]
      pi(nil) = []
      pi(inter) = [1, 2]

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: 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, r7, r15

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{2, x1 - 5}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 4
        false_A = 4
        app_A(x1,x2) = max{17, x1 + 15, x2 + 15}
        nil_A = 6
        inter_A(x1,x2) = max{x1 - 9, x2 + 16}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            eq#_A(x1,x2) = x1 + x2
            s_A(x1) = x1 + 1
            if_A(x1,x2,x3) = x2 + x3
            true_A() = 4
            false_A() = 4
            app_A(x1,x2) = x1 + x2 + 1
            nil_A() = 3
            inter_A(x1,x2) = x1 + 2
            eq#!_A(x1,x2) = x1 + x2 + 2
            s!_A(x1) = 1
            if!_A(x1,x2,x3) = x1 + x2 + x3
            true!_A() = 0
            false!_A() = 0
            app!_A(x1,x2) = x1 + 1
            nil!_A() = 1
            inter!_A(x1,x2) = x1 + 2
  
    partial status:
  
      pi(eq#) = []
      pi(s) = []
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(app) = []
      pi(nil) = []
      pi(inter) = [2]

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, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = 0
        cons_A(x1,x2) = max{19, x1 + 9}
        app_A(x1,x2) = max{28, x1 + 9, x2}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 6
        false_A = 2
        nil_A = 9
        inter_A(x1,x2) = max{x1 + 8, x2 + 8}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            app#_A(x1,x2) = x1 + x2
            cons_A(x1,x2) = x1 + x2 + 1
            app_A(x1,x2) = x1 + x2 + 1
            if_A(x1,x2,x3) = x2 + x3
            true_A() = 4
            false_A() = 6
            nil_A() = 8
            inter_A(x1,x2) = x1 + x2 + 1
            app#!_A(x1,x2) = x1 + x2 + 4
            cons!_A(x1,x2) = x1 + x2 + 3
            app!_A(x1,x2) = x1 + x2 + 6
            if!_A(x1,x2,x3) = x1 + x2 + x3
            true!_A() = 5
            false!_A() = 7
            nil!_A() = 8
            inter!_A(x1,x2) = x1 + x2 + 1
  
    partial status:
  
      pi(app#) = []
      pi(cons) = [1]
      pi(app) = [1, 2]
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(nil) = []
      pi(inter) = [1, 2]

The following rules are removed:

  p1, p2



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{0, x1 - 6}
        app_A(x1,x2) = max{5, x1 + 3, x2}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 8
        false_A = 7
        nil_A = 1
        cons_A(x1,x2) = max{7, x1 + 3}
        inter_A(x1,x2) = max{2, x2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            app#_A(x1,x2) = x1 + x2
            app_A(x1,x2) = x1 + x2 + 1
            if_A(x1,x2,x3) = x2 + x3
            true_A() = 4
            false_A() = 4
            nil_A() = 2
            cons_A(x1,x2) = x1 + 1
            inter_A(x1,x2) = x1 + 1
            app#!_A(x1,x2) = x1
            app!_A(x1,x2) = 2
            if!_A(x1,x2,x3) = x1 + x2 + x3
            true!_A() = 1
            false!_A() = 6
            nil!_A() = 2
            cons!_A(x1,x2) = x1 + x2 + 3
            inter!_A(x1,x2) = 2
  
    partial status:
  
      pi(app#) = []
      pi(app) = [1, 2]
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(nil) = []
      pi(cons) = [1]
      pi(inter) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.