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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          ifinter#_A(x1,x2,x3,x4) = x3 + x4 + 3
          false_A() = 0
          inter#_A(x1,x2) = x1 + x2 + 2
          cons_A(x1,x2) = x1 + x2 + 2
          mem_A(x1,x2) = x2
          true_A() = 0
          app_A(x1,x2) = x1 + x2 + 1
          eq_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 1
          s_A(x1) = x1 + 1
          ifmem_A(x1,x2,x3) = x3 + 1
          nil_A() = 1
  
    precedence:
    
      ifinter# = inter# = true = app = eq > cons = |0| = s = nil > mem = ifmem > false
    
    partial status:
  
      pi(ifinter#) = []
      pi(false) = []
      pi(inter#) = []
      pi(cons) = [2]
      pi(mem) = []
      pi(true) = []
      pi(app) = [2]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(ifmem) = []
      pi(nil) = []

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#(l1,cons(x,l2)) -> ifinter#(mem(x,l1),x,l2,l1)
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: inter#(l1,cons(x,l2)) -> ifinter#(mem(x,l1),x,l2,l1)
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,l2)
p6: inter#(l1,app(l2,l3)) -> inter#(l1,l3)
p7: inter#(cons(x,l1),l2) -> ifinter#(mem(x,l2),x,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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          inter#_A(x1,x2) = x1 + x2 + 2
          cons_A(x1,x2) = x2 + 2
          ifinter#_A(x1,x2,x3,x4) = x3 + x4 + 3
          mem_A(x1,x2) = x2 + 5
          true_A() = 1
          app_A(x1,x2) = x1 + x2 + 1
          eq_A(x1,x2) = x2 + 1
          |0|_A() = 2
          s_A(x1) = x1 + 2
          false_A() = 0
          ifmem_A(x1,x2,x3) = x3 + 6
          nil_A() = 0
  
    precedence:
    
      inter# = cons = ifinter# = true = eq > mem = app = |0| = ifmem = nil > s = false
    
    partial status:
  
      pi(inter#) = [2]
      pi(cons) = [2]
      pi(ifinter#) = []
      pi(mem) = []
      pi(true) = []
      pi(app) = [1, 2]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = []
      pi(false) = []
      pi(ifmem) = []
      pi(nil) = []

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#(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)

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          ifinter#_A(x1,x2,x3,x4) = x3 + x4 + 3
          true_A() = 1
          inter#_A(x1,x2) = x1 + x2 + 2
          cons_A(x1,x2) = x1 + x2 + 4
          mem_A(x1,x2) = x2 + 2
          app_A(x1,x2) = x1 + x2 + 1
          eq_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 4
          s_A(x1) = x1 + 2
          false_A() = 3
          ifmem_A(x1,x2,x3) = x3 + 3
          nil_A() = 2
  
    precedence:
    
      ifinter# = inter# = cons > true = mem = app = eq = |0| = s = false = ifmem = nil
    
    partial status:
  
      pi(ifinter#) = []
      pi(true) = []
      pi(inter#) = []
      pi(cons) = [2]
      pi(mem) = []
      pi(app) = [2]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = []
      pi(false) = []
      pi(ifmem) = []
      pi(nil) = []

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          ifinter#_A(x1,x2,x3,x4) = x3 + 3
          true_A() = 1
          inter#_A(x1,x2) = x1 + 2
          app_A(x1,x2) = x1 + x2 + 3
          cons_A(x1,x2) = x2 + 4
          mem_A(x1,x2) = x2 + 7
          eq_A(x1,x2) = 2
          |0|_A() = 2
          s_A(x1) = 3
          false_A() = 1
          ifmem_A(x1,x2,x3) = x3 + 8
          nil_A() = 0
  
    precedence:
    
      true = app > ifinter# = inter# = cons > mem = eq = |0| = s = false = ifmem = nil
    
    partial status:
  
      pi(ifinter#) = []
      pi(true) = []
      pi(inter#) = []
      pi(app) = [2]
      pi(cons) = [2]
      pi(mem) = [2]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = []
      pi(false) = []
      pi(ifmem) = [3]
      pi(nil) = []

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


-- Reduction pair.

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

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

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          ifinter#_A(x1,x2,x3,x4) = x3 + x4 + 6
          true_A() = 2
          inter#_A(x1,x2) = x1 + x2 + 3
          cons_A(x1,x2) = x2 + 5
          mem_A(x1,x2) = x2 + 3
          app_A(x1,x2) = x1 + x2 + 1
          eq_A(x1,x2) = 3
          |0|_A() = 1
          s_A(x1) = 4
          false_A() = 2
          ifmem_A(x1,x2,x3) = x3 + 4
          nil_A() = 0
  
    precedence:
    
      ifinter# = inter# > true = cons > mem = app = eq = |0| = s = false = ifmem = nil
    
    partial status:
  
      pi(ifinter#) = []
      pi(true) = []
      pi(inter#) = []
      pi(cons) = [2]
      pi(mem) = [2]
      pi(app) = [1, 2]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = []
      pi(false) = []
      pi(ifmem) = [3]
      pi(nil) = []

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

  {p2, p3}


-- Reduction pair.

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

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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          inter#_A(x1,x2) = x1 + x2 + 2
          app_A(x1,x2) = x1 + x2 + 1
  
    precedence:
    
      inter# = app
    
    partial status:
  
      pi(inter#) = []
      pi(app) = [2]

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#(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}


-- 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          inter#_A(x1,x2) = x1
          app_A(x1,x2) = x1 + x2 + 1
  
    precedence:
    
      inter# = app
    
    partial status:
  
      pi(inter#) = []
      pi(app) = [2]

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          ifmem#_A(x1,x2,x3) = x3 + 2
          false_A() = 2
          mem#_A(x1,x2) = x2 + 1
          cons_A(x1,x2) = x1 + x2 + 4
          eq_A(x1,x2) = 3
          |0|_A() = 3
          true_A() = 0
          s_A(x1) = x1 + 4
  
    precedence:
    
      cons > ifmem# > mem# > false = eq = |0| = true = s
    
    partial status:
  
      pi(ifmem#) = []
      pi(false) = []
      pi(mem#) = [2]
      pi(cons) = [1]
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          eq#_A(x1,x2) = x2
          s_A(x1) = x1 + 1
  
    precedence:
    
      eq# = s
    
    partial status:
  
      pi(eq#) = [2]
      pi(s) = [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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = x1 + x2 + 3
          cons_A(x1,x2) = x1 + x2 + 1
          app_A(x1,x2) = x1 + x2 + 2
          nil_A() = 1
  
    precedence:
    
      app# = cons = app = nil
    
    partial status:
  
      pi(app#) = []
      pi(cons) = [1, 2]
      pi(app) = [1]
      pi(nil) = []

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


-- Reduction pair.

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

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = x1 + x2 + 3
          app_A(x1,x2) = x1 + x2 + 2
          nil_A() = 1
          cons_A(x1,x2) = x1 + x2 + 1
  
    precedence:
    
      app# = app = nil = cons
    
    partial status:
  
      pi(app#) = [1]
      pi(app) = [1]
      pi(nil) = []
      pi(cons) = [1, 2]

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: 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

  r7, r8, r9

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = x1 + x2 + 1
          app_A(x1,x2) = x1 + x2 + 2
          nil_A() = 1
          cons_A(x1,x2) = x1 + x2 + 1
  
    precedence:
    
      app# = app = nil = cons
    
    partial status:
  
      pi(app#) = [1]
      pi(app) = [1]
      pi(nil) = []
      pi(cons) = [1, 2]

The next rules are strictly ordered:

  p1

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