YES

We show the termination of the TRS R:

  and(false(),false()) -> false()
  and(true(),false()) -> false()
  and(false(),true()) -> false()
  and(true(),true()) -> true()
  eq(nil(),nil()) -> true()
  eq(cons(T,L),nil()) -> false()
  eq(nil(),cons(T,L)) -> false()
  eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
  eq(var(L),var(Lp)) -> eq(L,Lp)
  eq(var(L),apply(T,S)) -> false()
  eq(var(L),lambda(X,T)) -> false()
  eq(apply(T,S),var(L)) -> false()
  eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
  eq(apply(T,S),lambda(X,Tp)) -> false()
  eq(lambda(X,T),var(L)) -> false()
  eq(lambda(X,T),apply(Tp,Sp)) -> false()
  eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
  if(true(),var(K),var(L)) -> var(K)
  if(false(),var(K),var(L)) -> var(L)
  ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
  ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
  ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

-- SCC decomposition.

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

p1: eq#(cons(T,L),cons(Tp,Lp)) -> and#(eq(T,Tp),eq(L,Lp))
p2: eq#(cons(T,L),cons(Tp,Lp)) -> eq#(T,Tp)
p3: eq#(cons(T,L),cons(Tp,Lp)) -> eq#(L,Lp)
p4: eq#(var(L),var(Lp)) -> eq#(L,Lp)
p5: eq#(apply(T,S),apply(Tp,Sp)) -> and#(eq(T,Tp),eq(S,Sp))
p6: eq#(apply(T,S),apply(Tp,Sp)) -> eq#(T,Tp)
p7: eq#(apply(T,S),apply(Tp,Sp)) -> eq#(S,Sp)
p8: eq#(lambda(X,T),lambda(Xp,Tp)) -> and#(eq(T,Tp),eq(X,Xp))
p9: eq#(lambda(X,T),lambda(Xp,Tp)) -> eq#(T,Tp)
p10: eq#(lambda(X,T),lambda(Xp,Tp)) -> eq#(X,Xp)
p11: ren#(var(L),var(K),var(Lp)) -> if#(eq(L,Lp),var(K),var(Lp))
p12: ren#(var(L),var(K),var(Lp)) -> eq#(L,Lp)
p13: ren#(X,Y,apply(T,S)) -> ren#(X,Y,T)
p14: ren#(X,Y,apply(T,S)) -> ren#(X,Y,S)
p15: ren#(X,Y,lambda(Z,T)) -> ren#(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T))
p16: ren#(X,Y,lambda(Z,T)) -> ren#(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)

and R consists of:

r1: and(false(),false()) -> false()
r2: and(true(),false()) -> false()
r3: and(false(),true()) -> false()
r4: and(true(),true()) -> true()
r5: eq(nil(),nil()) -> true()
r6: eq(cons(T,L),nil()) -> false()
r7: eq(nil(),cons(T,L)) -> false()
r8: eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
r9: eq(var(L),var(Lp)) -> eq(L,Lp)
r10: eq(var(L),apply(T,S)) -> false()
r11: eq(var(L),lambda(X,T)) -> false()
r12: eq(apply(T,S),var(L)) -> false()
r13: eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
r14: eq(apply(T,S),lambda(X,Tp)) -> false()
r15: eq(lambda(X,T),var(L)) -> false()
r16: eq(lambda(X,T),apply(Tp,Sp)) -> false()
r17: eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
r18: if(true(),var(K),var(L)) -> var(K)
r19: if(false(),var(K),var(L)) -> var(L)
r20: ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
r21: ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
r22: ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

The estimated dependency graph contains the following SCCs:

  {p13, p14, p15, p16}
  {p2, p3, p4, p6, p7, p9, p10}


-- Reduction pair.

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

p1: ren#(X,Y,lambda(Z,T)) -> ren#(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)
p2: ren#(X,Y,lambda(Z,T)) -> ren#(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T))
p3: ren#(X,Y,apply(T,S)) -> ren#(X,Y,S)
p4: ren#(X,Y,apply(T,S)) -> ren#(X,Y,T)

and R consists of:

r1: and(false(),false()) -> false()
r2: and(true(),false()) -> false()
r3: and(false(),true()) -> false()
r4: and(true(),true()) -> true()
r5: eq(nil(),nil()) -> true()
r6: eq(cons(T,L),nil()) -> false()
r7: eq(nil(),cons(T,L)) -> false()
r8: eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
r9: eq(var(L),var(Lp)) -> eq(L,Lp)
r10: eq(var(L),apply(T,S)) -> false()
r11: eq(var(L),lambda(X,T)) -> false()
r12: eq(apply(T,S),var(L)) -> false()
r13: eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
r14: eq(apply(T,S),lambda(X,Tp)) -> false()
r15: eq(lambda(X,T),var(L)) -> false()
r16: eq(lambda(X,T),apply(Tp,Sp)) -> false()
r17: eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
r18: if(true(),var(K),var(L)) -> var(K)
r19: if(false(),var(K),var(L)) -> var(L)
r20: ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
r21: ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
r22: ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ren#_A(x1,x2,x3) = x3 + 58
          lambda_A(x1,x2) = max{x1 + 109, x2 + 59}
          var_A(x1) = 0
          cons_A(x1,x2) = max{33, x1 - 121, x2 - 47}
          nil_A = 0
          ren_A(x1,x2,x3) = max{35, x3}
          apply_A(x1,x2) = max{228, x1 + 159, x2 + 193}
          and_A(x1,x2) = max{277, x2}
          false_A = 0
          true_A = 0
          eq_A(x1,x2) = 278
          if_A(x1,x2,x3) = max{x1 - 279, x2 + 35, x3 - 4}
    
      precedence:
      
        nil > ren = if > true > apply > var > ren# = lambda = cons = and = eq > false
      
      partial status:
    
        pi(ren#) = [3]
        pi(lambda) = [1, 2]
        pi(var) = []
        pi(cons) = []
        pi(nil) = []
        pi(ren) = [3]
        pi(apply) = [1, 2]
        pi(and) = []
        pi(false) = []
        pi(true) = []
        pi(eq) = []
        pi(if) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ren#_A(x1,x2,x3) = max{25, x3 - 6}
          lambda_A(x1,x2) = max{x1 + 7, x2}
          var_A(x1) = 17
          cons_A(x1,x2) = 92
          nil_A = 176
          ren_A(x1,x2,x3) = max{24, x3 - 6}
          apply_A(x1,x2) = max{24, x1, x2}
          and_A(x1,x2) = 52
          false_A = 4
          true_A = 0
          eq_A(x1,x2) = 53
          if_A(x1,x2,x3) = 24
    
      precedence:
      
        cons = ren > ren# = var > lambda = apply = and = false = true = eq > nil = if
      
      partial status:
    
        pi(ren#) = []
        pi(lambda) = [1]
        pi(var) = []
        pi(cons) = []
        pi(nil) = []
        pi(ren) = []
        pi(apply) = []
        pi(and) = []
        pi(false) = []
        pi(true) = []
        pi(eq) = []
        pi(if) = []
    

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: ren#(X,Y,lambda(Z,T)) -> ren#(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)
p2: ren#(X,Y,lambda(Z,T)) -> ren#(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T))
p3: ren#(X,Y,apply(T,S)) -> ren#(X,Y,S)

and R consists of:

r1: and(false(),false()) -> false()
r2: and(true(),false()) -> false()
r3: and(false(),true()) -> false()
r4: and(true(),true()) -> true()
r5: eq(nil(),nil()) -> true()
r6: eq(cons(T,L),nil()) -> false()
r7: eq(nil(),cons(T,L)) -> false()
r8: eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
r9: eq(var(L),var(Lp)) -> eq(L,Lp)
r10: eq(var(L),apply(T,S)) -> false()
r11: eq(var(L),lambda(X,T)) -> false()
r12: eq(apply(T,S),var(L)) -> false()
r13: eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
r14: eq(apply(T,S),lambda(X,Tp)) -> false()
r15: eq(lambda(X,T),var(L)) -> false()
r16: eq(lambda(X,T),apply(Tp,Sp)) -> false()
r17: eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
r18: if(true(),var(K),var(L)) -> var(K)
r19: if(false(),var(K),var(L)) -> var(L)
r20: ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
r21: ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
r22: ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: ren#(X,Y,lambda(Z,T)) -> ren#(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)
p2: ren#(X,Y,apply(T,S)) -> ren#(X,Y,S)
p3: ren#(X,Y,lambda(Z,T)) -> ren#(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T))

and R consists of:

r1: and(false(),false()) -> false()
r2: and(true(),false()) -> false()
r3: and(false(),true()) -> false()
r4: and(true(),true()) -> true()
r5: eq(nil(),nil()) -> true()
r6: eq(cons(T,L),nil()) -> false()
r7: eq(nil(),cons(T,L)) -> false()
r8: eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
r9: eq(var(L),var(Lp)) -> eq(L,Lp)
r10: eq(var(L),apply(T,S)) -> false()
r11: eq(var(L),lambda(X,T)) -> false()
r12: eq(apply(T,S),var(L)) -> false()
r13: eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
r14: eq(apply(T,S),lambda(X,Tp)) -> false()
r15: eq(lambda(X,T),var(L)) -> false()
r16: eq(lambda(X,T),apply(Tp,Sp)) -> false()
r17: eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
r18: if(true(),var(K),var(L)) -> var(K)
r19: if(false(),var(K),var(L)) -> var(L)
r20: ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
r21: ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
r22: ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ren#_A(x1,x2,x3) = max{5, x3 - 103}
          lambda_A(x1,x2) = x2 + 145
          var_A(x1) = 0
          cons_A(x1,x2) = 2
          nil_A = 144
          apply_A(x1,x2) = max{x1 + 14, x2 + 109}
          ren_A(x1,x2,x3) = x3
          and_A(x1,x2) = max{145, x1 - 92, x2 - 7}
          false_A = 52
          true_A = 145
          eq_A(x1,x2) = 154
          if_A(x1,x2,x3) = max{x1 - 154, x2}
    
      precedence:
      
        ren# = cons = ren > and > lambda > apply = false > nil > eq > true = if > var
      
      partial status:
    
        pi(ren#) = []
        pi(lambda) = [2]
        pi(var) = []
        pi(cons) = []
        pi(nil) = []
        pi(apply) = [1, 2]
        pi(ren) = []
        pi(and) = []
        pi(false) = []
        pi(true) = []
        pi(eq) = []
        pi(if) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ren#_A(x1,x2,x3) = 88
          lambda_A(x1,x2) = 87
          var_A(x1) = 86
          cons_A(x1,x2) = 215
          nil_A = 48
          apply_A(x1,x2) = 87
          ren_A(x1,x2,x3) = 87
          and_A(x1,x2) = 218
          false_A = 111
          true_A = 51
          eq_A(x1,x2) = 159
          if_A(x1,x2,x3) = 86
    
      precedence:
      
        ren = false > cons = apply > lambda > ren# = nil > and > true = eq = if > var
      
      partial status:
    
        pi(ren#) = []
        pi(lambda) = []
        pi(var) = []
        pi(cons) = []
        pi(nil) = []
        pi(apply) = []
        pi(ren) = []
        pi(and) = []
        pi(false) = []
        pi(true) = []
        pi(eq) = []
        pi(if) = []
    

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: ren#(X,Y,lambda(Z,T)) -> ren#(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)
p2: ren#(X,Y,lambda(Z,T)) -> ren#(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T))

and R consists of:

r1: and(false(),false()) -> false()
r2: and(true(),false()) -> false()
r3: and(false(),true()) -> false()
r4: and(true(),true()) -> true()
r5: eq(nil(),nil()) -> true()
r6: eq(cons(T,L),nil()) -> false()
r7: eq(nil(),cons(T,L)) -> false()
r8: eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
r9: eq(var(L),var(Lp)) -> eq(L,Lp)
r10: eq(var(L),apply(T,S)) -> false()
r11: eq(var(L),lambda(X,T)) -> false()
r12: eq(apply(T,S),var(L)) -> false()
r13: eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
r14: eq(apply(T,S),lambda(X,Tp)) -> false()
r15: eq(lambda(X,T),var(L)) -> false()
r16: eq(lambda(X,T),apply(Tp,Sp)) -> false()
r17: eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
r18: if(true(),var(K),var(L)) -> var(K)
r19: if(false(),var(K),var(L)) -> var(L)
r20: ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
r21: ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
r22: ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: ren#(X,Y,lambda(Z,T)) -> ren#(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)
p2: ren#(X,Y,lambda(Z,T)) -> ren#(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T))

and R consists of:

r1: and(false(),false()) -> false()
r2: and(true(),false()) -> false()
r3: and(false(),true()) -> false()
r4: and(true(),true()) -> true()
r5: eq(nil(),nil()) -> true()
r6: eq(cons(T,L),nil()) -> false()
r7: eq(nil(),cons(T,L)) -> false()
r8: eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
r9: eq(var(L),var(Lp)) -> eq(L,Lp)
r10: eq(var(L),apply(T,S)) -> false()
r11: eq(var(L),lambda(X,T)) -> false()
r12: eq(apply(T,S),var(L)) -> false()
r13: eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
r14: eq(apply(T,S),lambda(X,Tp)) -> false()
r15: eq(lambda(X,T),var(L)) -> false()
r16: eq(lambda(X,T),apply(Tp,Sp)) -> false()
r17: eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
r18: if(true(),var(K),var(L)) -> var(K)
r19: if(false(),var(K),var(L)) -> var(L)
r20: ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
r21: ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
r22: ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ren#_A(x1,x2,x3) = x3 + 3
          lambda_A(x1,x2) = max{119, x1 + 55, x2 + 50}
          var_A(x1) = 59
          cons_A(x1,x2) = x2 + 28
          nil_A = 26
          ren_A(x1,x2,x3) = x3
          and_A(x1,x2) = max{58, x1 - 72, x2 - 1}
          false_A = 0
          true_A = 25
          eq_A(x1,x2) = 58
          apply_A(x1,x2) = max{x1 + 28, x2 + 45}
          if_A(x1,x2,x3) = max{58, x1 - 11, x2}
    
      precedence:
      
        ren# = cons = nil = true > ren > lambda = false = eq > var = and = apply = if
      
      partial status:
    
        pi(ren#) = [3]
        pi(lambda) = [1, 2]
        pi(var) = []
        pi(cons) = [2]
        pi(nil) = []
        pi(ren) = []
        pi(and) = []
        pi(false) = []
        pi(true) = []
        pi(eq) = []
        pi(apply) = [2]
        pi(if) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ren#_A(x1,x2,x3) = x3 + 98
          lambda_A(x1,x2) = max{x1 - 26, x2 + 14}
          var_A(x1) = 28
          cons_A(x1,x2) = max{37, x2 + 17}
          nil_A = 10
          ren_A(x1,x2,x3) = 3
          and_A(x1,x2) = 21
          false_A = 11
          true_A = 22
          eq_A(x1,x2) = 21
          apply_A(x1,x2) = x2 + 12
          if_A(x1,x2,x3) = max{18, x2}
    
      precedence:
      
        ren# = lambda = var = cons = nil = ren = and = false = true = eq = apply = if
      
      partial status:
    
        pi(ren#) = []
        pi(lambda) = []
        pi(var) = []
        pi(cons) = [2]
        pi(nil) = []
        pi(ren) = []
        pi(and) = []
        pi(false) = []
        pi(true) = []
        pi(eq) = []
        pi(apply) = [2]
        pi(if) = [2]
    

The next rules are strictly ordered:

  p1, p2

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: eq#(cons(T,L),cons(Tp,Lp)) -> eq#(T,Tp)
p2: eq#(lambda(X,T),lambda(Xp,Tp)) -> eq#(X,Xp)
p3: eq#(lambda(X,T),lambda(Xp,Tp)) -> eq#(T,Tp)
p4: eq#(apply(T,S),apply(Tp,Sp)) -> eq#(S,Sp)
p5: eq#(apply(T,S),apply(Tp,Sp)) -> eq#(T,Tp)
p6: eq#(var(L),var(Lp)) -> eq#(L,Lp)
p7: eq#(cons(T,L),cons(Tp,Lp)) -> eq#(L,Lp)

and R consists of:

r1: and(false(),false()) -> false()
r2: and(true(),false()) -> false()
r3: and(false(),true()) -> false()
r4: and(true(),true()) -> true()
r5: eq(nil(),nil()) -> true()
r6: eq(cons(T,L),nil()) -> false()
r7: eq(nil(),cons(T,L)) -> false()
r8: eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
r9: eq(var(L),var(Lp)) -> eq(L,Lp)
r10: eq(var(L),apply(T,S)) -> false()
r11: eq(var(L),lambda(X,T)) -> false()
r12: eq(apply(T,S),var(L)) -> false()
r13: eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
r14: eq(apply(T,S),lambda(X,Tp)) -> false()
r15: eq(lambda(X,T),var(L)) -> false()
r16: eq(lambda(X,T),apply(Tp,Sp)) -> false()
r17: eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
r18: if(true(),var(K),var(L)) -> var(K)
r19: if(false(),var(K),var(L)) -> var(L)
r20: ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
r21: ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
r22: ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          eq#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
          cons_A(x1,x2) = max{x1, x2}
          lambda_A(x1,x2) = max{x1, x2}
          apply_A(x1,x2) = max{x1, x2}
          var_A(x1) = max{1, x1}
    
      precedence:
      
        eq# = cons = lambda = apply = var
      
      partial status:
    
        pi(eq#) = [1, 2]
        pi(cons) = [1, 2]
        pi(lambda) = [1, 2]
        pi(apply) = [1, 2]
        pi(var) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          eq#_A(x1,x2) = max{x1 - 1, x2 + 1}
          cons_A(x1,x2) = max{x1, x2}
          lambda_A(x1,x2) = max{x1, x2}
          apply_A(x1,x2) = max{x1, x2}
          var_A(x1) = x1
    
      precedence:
      
        eq# = cons = lambda = apply = var
      
      partial status:
    
        pi(eq#) = [2]
        pi(cons) = [1, 2]
        pi(lambda) = [1, 2]
        pi(apply) = [1, 2]
        pi(var) = [1]
    

The next rules are strictly ordered:

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

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