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, r12, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ifinter#_A(x1,x2,x3,x4) = max{x2 + 22, x3 + 37, x4 + 35}
        false_A = 20
        inter#_A(x1,x2) = max{x1 + 36, x2 + 35}
        cons_A(x1,x2) = max{14, x1 - 12, x2 + 7}
        mem_A(x1,x2) = max{19, x1 + 6, x2 + 4}
        true_A = 13
        app_A(x1,x2) = max{4, x1, x2}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        eq_A(x1,x2) = max{30, x1 + 9, x2 - 2}
        |0|_A = 3
        s_A(x1) = max{21, x1 - 10}
        ifmem_A(x1,x2,x3) = max{20, x1 - 2, x2 + 6, x3 + 12}
        inter_A(x1,x2) = max{12, x1 + 3, x2 + 1}
        nil_A = 13
  
    base order 2:
  
      precedence:
      
        false = eq = ifmem > cons = s > true > ifinter# = inter# = if = |0| > mem > app = inter > nil
      
  
    partial status:
  
      pi(ifinter#) = []
      pi(false) = []
      pi(inter#) = []
      pi(cons) = [2]
      pi(mem) = []
      pi(true) = []
      pi(app) = [1, 2]
      pi(if) = [2, 3]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = []
      pi(ifmem) = [3]
      pi(inter) = [2]
      pi(nil) = []

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, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ifinter#_A(x1,x2,x3,x4) = x4 + 30
        false_A = 27
        inter#_A(x1,x2) = x2 + 30
        app_A(x1,x2) = max{55, x1 + 31, x2 + 32}
        cons_A(x1,x2) = max{x1, x2 - 22}
        mem_A(x1,x2) = max{x1 - 12, x2 + 26}
        true_A = 6
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        eq_A(x1,x2) = max{3, x1 - 23, x2 + 2}
        |0|_A = 28
        s_A(x1) = x1 + 27
        nil_A = 0
        inter_A(x1,x2) = max{21, x1 - 20, x2 - 11}
  
    base order 2:
  
      precedence:
      
        |0| > true > ifinter# = inter# = app = if = inter > cons = nil > mem > eq > s > false
      
  
    partial status:
  
      pi(ifinter#) = []
      pi(false) = []
      pi(inter#) = []
      pi(app) = [1, 2]
      pi(cons) = [1]
      pi(mem) = [2]
      pi(true) = []
      pi(if) = [2, 3]
      pi(eq) = [2]
      pi(|0|) = []
      pi(s) = [1]
      pi(nil) = []
      pi(inter) = []

The following rules are removed:

  p5



-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

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

  r1, r2, r3, r4, r5, r10, r12, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ifinter#_A(x1,x2,x3,x4) = x4 + 24
        false_A = 1
        inter#_A(x1,x2) = x2 + 24
        cons_A(x1,x2) = max{2, x1 - 26, x2 - 14}
        mem_A(x1,x2) = max{3, x1 - 45, x2 - 45}
        true_A = 1
        app_A(x1,x2) = max{x1 + 25, x2 - 19}
        if_A(x1,x2,x3) = max{x1, x2 + 1, x3 + 1}
        eq_A(x1,x2) = max{3, x1 - 45, x2 - 21}
        |0|_A = 2
        s_A(x1) = max{2, x1 - 46}
        nil_A = 2
        ifmem_A(x1,x2,x3) = max{3, x1 - 4, x2 - 45, x3 - 45}
        inter_A(x1,x2) = max{13, x1 - 18, x2 - 18}
  
    base order 2:
  
      precedence:
      
        app = if = |0| = ifmem = inter > mem = nil > eq > true = s > ifinter# = inter# > false = cons
      
  
    partial status:
  
      pi(ifinter#) = []
      pi(false) = []
      pi(inter#) = []
      pi(cons) = []
      pi(mem) = []
      pi(true) = []
      pi(app) = [1]
      pi(if) = [1, 2, 3]
      pi(eq) = []
      pi(|0|) = []
      pi(s) = []
      pi(nil) = []
      pi(ifmem) = []
      pi(inter) = []

The following rules are removed:

  p4



-- SCC decomposition.

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

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

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ifinter#_A(x1,x2,x3,x4) = max{x2 + 26, x3 + 16}
        false_A = 25
        inter#_A(x1,x2) = max{15, x1 - 1}
        app_A(x1,x2) = max{37, x1 + 18, x2 + 17}
        cons_A(x1,x2) = max{37, x1 + 28, x2 + 18}
        mem_A(x1,x2) = max{14, x1 + 9, x2 - 27}
        true_A = 12
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        eq_A(x1,x2) = max{49, x1 + 22, x2 - 1}
        |0|_A = 4
        s_A(x1) = x1 + 2
  
    base order 2:
  
      precedence:
      
        eq > true > inter# > false > mem = if = |0| = s > app = cons > ifinter#
      
  
    partial status:
  
      pi(ifinter#) = []
      pi(false) = []
      pi(inter#) = []
      pi(app) = [1, 2]
      pi(cons) = [2]
      pi(mem) = [1]
      pi(true) = []
      pi(if) = [2, 3]
      pi(eq) = [1]
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p4



-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p3}


-- 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, r3, r4, r5, r7, r10, r12, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        inter#_A(x1,x2) = max{8, x1 - 1}
        app_A(x1,x2) = max{x1 + 10, x2 + 5}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 1
        false_A = 1
        eq_A(x1,x2) = max{10, x1 - 3, x2 - 23}
        |0|_A = 2
        s_A(x1) = max{2, x1 - 24}
        nil_A = 2
        mem_A(x1,x2) = max{9, x1 - 15, x2 - 10}
        ifmem_A(x1,x2,x3) = max{10, x1 - 11, x2 - 1, x3 - 1}
        inter_A(x1,x2) = max{6, x1 - 29, x2 - 29}
  
    base order 2:
  
      precedence:
      
        if = eq = |0| = ifmem = inter > app = s = nil = mem > inter# = false > true
      
  
    partial status:
  
      pi(inter#) = []
      pi(app) = [2]
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(eq) = []
      pi(|0|) = []
      pi(s) = []
      pi(nil) = []
      pi(mem) = []
      pi(ifmem) = []
      pi(inter) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: 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}


-- Reduction pair.

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

p1: 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, r10, r12, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        inter#_A(x1,x2) = x1 + 8
        app_A(x1,x2) = max{x1 + 12, x2 + 1}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 1
        false_A = 1
        eq_A(x1,x2) = max{3, x1 - 28, x2 - 28}
        |0|_A = 2
        s_A(x1) = max{2, x1 - 29}
        mem_A(x1,x2) = max{4, x1 - 27, x2 - 10}
        nil_A = 2
        ifmem_A(x1,x2,x3) = max{4, x1 - 4, x2 - 27, x3 - 10}
        inter_A(x1,x2) = max{5, x1 - 9, x2 - 8}
  
    base order 2:
  
      precedence:
      
        if = |0| = ifmem = inter > app = mem = nil > inter# = eq > s > true = false
      
  
    partial status:
  
      pi(inter#) = [1]
      pi(app) = [1]
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(eq) = []
      pi(|0|) = []
      pi(s) = []
      pi(mem) = []
      pi(nil) = []
      pi(ifmem) = []
      pi(inter) = []

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: 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, r10, r12, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ifmem#_A(x1,x2,x3) = max{3, x3 - 2}
        false_A = 1
        mem#_A(x1,x2) = max{2, x2 - 3}
        cons_A(x1,x2) = max{7, x1 - 2, x2 + 2}
        eq_A(x1,x2) = max{4, x1 - 25, x2 - 25}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 1
        |0|_A = 24
        s_A(x1) = max{2, x1 - 26}
        mem_A(x1,x2) = max{8, x1 - 24, x2 - 20}
        nil_A = 2
        ifmem_A(x1,x2,x3) = max{9, x1 - 27, x2 - 23, x3 - 17}
        inter_A(x1,x2) = max{19, x1 - 19, x2 - 10}
  
    base order 2:
  
      precedence:
      
        ifmem = inter > mem# = mem > ifmem# = eq = if = nil > cons = |0| = s > false = true
      
  
    partial status:
  
      pi(ifmem#) = []
      pi(false) = []
      pi(mem#) = []
      pi(cons) = [2]
      pi(eq) = []
      pi(if) = [2, 3]
      pi(true) = []
      pi(|0|) = []
      pi(s) = []
      pi(mem) = []
      pi(nil) = []
      pi(ifmem) = []
      pi(inter) = []

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, r3, r4, r5, r10, r12, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        eq#_A(x1,x2) = max{4, x1 + 2}
        s_A(x1) = max{67, x1 + 3}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 6
        false_A = 4
        eq_A(x1,x2) = max{18, x1 - 68, x2 - 48}
        |0|_A = 53
        mem_A(x1,x2) = max{19, x1 - 2, x2 - 19}
        nil_A = 14
        ifmem_A(x1,x2,x3) = max{20, x1 - 19, x2 - 1, x3 - 18}
        inter_A(x1,x2) = max{15, x1 - 40, x2 - 28}
  
    base order 2:
  
      precedence:
      
        ifmem > mem > eq > eq# = s = if = |0| = inter > true = nil > false
      
  
    partial status:
  
      pi(eq#) = [1]
      pi(s) = [1]
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(eq) = []
      pi(|0|) = []
      pi(mem) = []
      pi(nil) = []
      pi(ifmem) = []
      pi(inter) = []

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, r3, r4, r5, r7, r10, r12, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = x1 + 23
        cons_A(x1,x2) = max{x1 - 11, x2 + 24}
        app_A(x1,x2) = max{x1 + 10, x2 + 5}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 5
        false_A = 1
        eq_A(x1,x2) = max{x1 - 7, x2 + 6}
        |0|_A = 6
        s_A(x1) = max{2, x1 - 8}
        nil_A = 10
        mem_A(x1,x2) = max{44, x1 - 1, x2 + 18}
        ifmem_A(x1,x2,x3) = max{8, x1 - 16, x2 - 2, x3 - 1}
        inter_A(x1,x2) = max{10, x2 - 26}
  
    base order 2:
  
      precedence:
      
        mem > app# = app = eq = ifmem = inter > cons = if = |0| = s = nil > true = false
      
  
    partial status:
  
      pi(app#) = []
      pi(cons) = [2]
      pi(app) = []
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(eq) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(nil) = []
      pi(mem) = []
      pi(ifmem) = []
      pi(inter) = []

The following rules are removed:

  p2



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

  r1, r2, r3, r4, r5, r7, r10, r12, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = x1 + 5
        cons_A(x1,x2) = max{8, x1 - 3, x2 + 6}
        app_A(x1,x2) = max{11, x1 + 7, x2 + 7}
        if_A(x1,x2,x3) = max{x1, x2 + 1, x3 + 1}
        true_A = 32
        false_A = 1
        eq_A(x1,x2) = max{33, x1 - 34, x2 - 34}
        |0|_A = 33
        s_A(x1) = max{32, x1 - 33}
        nil_A = 2
        mem_A(x1,x2) = max{11, x1 - 2, x2 - 66}
        ifmem_A(x1,x2,x3) = max{33, x1 - 34, x2 - 2, x3 - 59}
        inter_A(x1,x2) = max{10, x1 - 82, x2 - 82}
  
    base order 2:
  
      precedence:
      
        ifmem > if = mem > eq = |0| > true > inter > cons = app = nil > app# = false > s
      
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = []
      pi(app) = [2]
      pi(if) = [1, 2, 3]
      pi(true) = []
      pi(false) = []
      pi(eq) = []
      pi(|0|) = []
      pi(s) = []
      pi(nil) = []
      pi(mem) = []
      pi(ifmem) = []
      pi(inter) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10, r12, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{5, x1 + 1}
        cons_A(x1,x2) = max{x1 + 8, x2 + 8}
        if_A(x1,x2,x3) = max{x2 + 1, x3 + 1}
        true_A = 2
        false_A = 2
        eq_A(x1,x2) = max{9, x1 - 6, x2 - 4}
        |0|_A = 3
        s_A(x1) = max{5, x1 - 7}
        mem_A(x1,x2) = max{10, x1 - 2, x2 - 10}
        nil_A = 3
        ifmem_A(x1,x2,x3) = max{9, x1 - 10, x2 - 3, x3 - 11}
        inter_A(x1,x2) = max{4, x1 - 22}
  
    base order 2:
  
      precedence:
      
        mem > eq = |0| = ifmem > true > cons = if = s > app# = inter > nil > false
      
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = []
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(eq) = []
      pi(|0|) = []
      pi(s) = []
      pi(mem) = []
      pi(nil) = []
      pi(ifmem) = []
      pi(inter) = []

The following rules are removed:

  p1

Then no dependency pair remains.