We consider the system Applicative_first_order_05__#3.48.

  Alphabet:

    0 : [] --> b 
    1 : [] --> b 
    c : [b] --> b 
    cons : [c * d] --> d 
    f : [b] --> a 
    false : [] --> a 
    filter : [c -> a * d] --> d 
    filter2 : [a * c -> a * c * d] --> d 
    g : [b * b] --> b 
    if : [a * b * b] --> b 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    s : [b] --> b 
    true : [] --> a 

  Rules:

    f(0) => true 
    f(1) => false 
    f(s(x)) => f(x) 
    if(true, s(x), s(y)) => s(x) 
    if(false, s(x), s(y)) => s(y) 
    g(x, c(y)) => c(g(x, y)) 
    g(x, c(y)) => g(x, if(f(x), c(g(s(x), y)), c(y))) 
    map(h, nil) => nil 
    map(h, cons(x, y)) => cons(h x, map(h, y)) 
    filter(h, nil) => nil 
    filter(h, cons(x, y)) => filter2(h x, h, x, y) 
    filter2(true, h, x, y) => cons(x, filter(h, y)) 
    filter2(false, h, x, y) => filter(h, y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We observe that the rules contain a first-order subset:

  f(0) => true 
  f(1) => false 
  f(s(X)) => f(X) 
  if(true, s(X), s(Y)) => s(X) 
  if(false, s(X), s(Y)) => s(Y) 
  g(X, c(Y)) => c(g(X, Y)) 
  g(X, c(Y)) => g(X, if(f(X), c(g(s(X), Y)), c(Y))) 

Moreover, the system is finitely branching.  Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is Ce-terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed Ce-terminating:

 || proof of resources/system.trs
 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
 || 
 || 
 || Termination w.r.t. Q of the given QTRS could be proven:
 || 
 || (0) QTRS
 || (1) DependencyPairsProof [EQUIVALENT]
 || (2) QDP
 || (3) DependencyGraphProof [EQUIVALENT]
 || (4) AND
 ||     (5) QDP
 ||         (6) UsableRulesProof [EQUIVALENT]
 ||         (7) QDP
 ||         (8) QDPSizeChangeProof [EQUIVALENT]
 ||         (9) YES
 ||     (10) QDP
 ||         (11) UsableRulesProof [EQUIVALENT]
 ||         (12) QDP
 ||         (13) QDPSizeChangeProof [EQUIVALENT]
 ||         (14) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    f(0) -> true
 ||    f(1) -> false
 ||    f(s(%X)) -> f(%X)
 ||    if(true, s(%X), s(%Y)) -> s(%X)
 ||    if(false, s(%X), s(%Y)) -> s(%Y)
 ||    g(%X, c(%Y)) -> c(g(%X, %Y))
 ||    g(%X, c(%Y)) -> g(%X, if(f(%X), c(g(s(%X), %Y)), c(%Y)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) DependencyPairsProof (EQUIVALENT)
 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    F(s(%X)) -> F(%X)
 ||    G(%X, c(%Y)) -> G(%X, %Y)
 ||    G(%X, c(%Y)) -> G(%X, if(f(%X), c(g(s(%X), %Y)), c(%Y)))
 ||    G(%X, c(%Y)) -> IF(f(%X), c(g(s(%X), %Y)), c(%Y))
 ||    G(%X, c(%Y)) -> F(%X)
 ||    G(%X, c(%Y)) -> G(s(%X), %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    f(0) -> true
 ||    f(1) -> false
 ||    f(s(%X)) -> f(%X)
 ||    if(true, s(%X), s(%Y)) -> s(%X)
 ||    if(false, s(%X), s(%Y)) -> s(%Y)
 ||    g(%X, c(%Y)) -> c(g(%X, %Y))
 ||    g(%X, c(%Y)) -> g(%X, if(f(%X), c(g(s(%X), %Y)), c(%Y)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (3) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
 || ----------------------------------------
 || 
 || (4)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (5)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    F(s(%X)) -> F(%X)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    f(0) -> true
 ||    f(1) -> false
 ||    f(s(%X)) -> f(%X)
 ||    if(true, s(%X), s(%Y)) -> s(%X)
 ||    if(false, s(%X), s(%Y)) -> s(%Y)
 ||    g(%X, c(%Y)) -> c(g(%X, %Y))
 ||    g(%X, c(%Y)) -> g(%X, if(f(%X), c(g(s(%X), %Y)), c(%Y)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (6) UsableRulesProof (EQUIVALENT)
 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (7)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    F(s(%X)) -> F(%X)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (8) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *F(s(%X)) -> F(%X)
 || The graph contains the following edges 1 > 1
 || 
 || 
 || ----------------------------------------
 || 
 || (9)
 || YES
 || 
 || ----------------------------------------
 || 
 || (10)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    G(%X, c(%Y)) -> G(s(%X), %Y)
 ||    G(%X, c(%Y)) -> G(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    f(0) -> true
 ||    f(1) -> false
 ||    f(s(%X)) -> f(%X)
 ||    if(true, s(%X), s(%Y)) -> s(%X)
 ||    if(false, s(%X), s(%Y)) -> s(%Y)
 ||    g(%X, c(%Y)) -> c(g(%X, %Y))
 ||    g(%X, c(%Y)) -> g(%X, if(f(%X), c(g(s(%X), %Y)), c(%Y)))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (11) UsableRulesProof (EQUIVALENT)
 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (12)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    G(%X, c(%Y)) -> G(s(%X), %Y)
 ||    G(%X, c(%Y)) -> G(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (13) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *G(%X, c(%Y)) -> G(s(%X), %Y)
 || The graph contains the following edges 2 > 2
 || 
 || 
 || *G(%X, c(%Y)) -> G(%X, %Y)
 || The graph contains the following edges 1 >= 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (14)
 || YES
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] map#(F, cons(X, Y)) =#> F(X)   
    1] map#(F, cons(X, Y)) =#> map#(F, Y)   
    2] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    3] filter#(F, cons(X, Y)) =#> F(X)   
    4] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    5] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    f(0) => true 
    f(1) => false 
    f(s(X)) => f(X) 
    if(true, s(X), s(Y)) => s(X) 
    if(false, s(X), s(Y)) => s(Y) 
    g(X, c(Y)) => c(g(X, Y)) 
    g(X, c(Y)) => g(X, if(f(X), c(g(s(X), Y)), c(Y))) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(F, Y) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

The formative rules of (P_0, R_0) are R_1 ::=

  f(0) => true 
  f(1) => false 
  f(s(X)) => f(X) 
  if(true, s(X), s(Y)) => s(X) 
  if(false, s(X), s(Y)) => s(Y) 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) => filter(F, Y) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative).

Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_1, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

  map#(F, cons(X, Y)) >? F(X) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter#(F, cons(X, Y)) >? F(X) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  filter2#(false, F, X, Y) >? filter#(F, Y) 
  f(0) >= true 
  f(1) >= false 
  f(s(X)) >= f(X) 
  if(true, s(X), s(Y)) >= s(X) 
  if(false, s(X), s(Y)) >= s(Y) 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >= filter(F, Y) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 3 
  1 = 3 
  cons = \y0y1.1 + y0 + y1 
  f = \y0.3 
  false = 0 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.1 + y2 + y3 
  filter2# = \y0G1y2y3.G1(y3) 
  filter# = \G0y1.G0(y1) 
  if = \y0y1y2.3 + y2 + 2y1 
  map = \G0y1.2y1 + G0(y1) + 2y1G0(y1) 
  map# = \G0y1.3 + y1 + G0(y1) 
  s = \y0.y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(1 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(1 + x1 + x2) > 3 + x2 + F0(x2) = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = F0(1 + x1 + x2) >= F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = F0(1 + x1 + x2) >= F0(x1) = [[_F0(_x1)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = F0(x2) >= F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = F0(x2) >= F0(x2) = [[filter#(_F0, _x2)]] 
  [[f(0)]] = 3 >= 0 = [[true]] 
  [[f(1)]] = 3 >= 0 = [[false]] 
  [[f(s(_x0))]] = 3 >= 3 = [[f(_x0)]] 
  [[if(true, s(_x0), s(_x1))]] = 3 + x1 + 2x0 >= x0 = [[s(_x0)]] 
  [[if(false, s(_x0), s(_x1))]] = 3 + x1 + 2x0 >= x1 = [[s(_x1)]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) >= 1 + 2x2 + F0(x2) + 2x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 1 + x1 + x2 >= x2 = [[filter(_F0, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_1, minimal, formative) by (P_1, R_1, minimal, formative), where P_1 consists of:

  filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
  filter#(F, cons(X, Y)) =#> F(X)   
  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_1, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter#(F, cons(X, Y)) >? F(X) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  filter2#(false, F, X, Y) >? filter#(F, Y) 
  f(0) >= true 
  f(1) >= false 
  f(s(X)) >= f(X) 
  if(true, s(X), s(Y)) >= s(X) 
  if(false, s(X), s(Y)) >= s(Y) 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >= filter(F, Y) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 3 
  1 = 3 
  cons = \y0y1.2 + y0 + y1 
  f = \y0.3 
  false = 0 
  filter = \G0y1.y1 + G0(y1) + y1G0(y1) 
  filter2 = \y0G1y2y3.2 + y2 + y3 + 2G1(y3) + y3G1(y3) 
  filter2# = \y0G1y2y3.2y3 + 2G1(y3) 
  filter# = \G0y1.2y1 + 2G0(y1) 
  if = \y0y1y2.3 
  map = \G0y1.y1 + y1G0(y1) 
  s = \y0.0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + 2F0(2 + x1 + x2) > 2x2 + 2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + 2F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 2x2 + 2F0(x2) >= 2x2 + 2F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 2x2 + 2F0(x2) >= 2x2 + 2F0(x2) = [[filter#(_F0, _x2)]] 
  [[f(0)]] = 3 >= 0 = [[true]] 
  [[f(1)]] = 3 >= 0 = [[false]] 
  [[f(s(_x0))]] = 3 >= 3 = [[f(_x0)]] 
  [[if(true, s(_x0), s(_x1))]] = 3 >= 0 = [[s(_x0)]] 
  [[if(false, s(_x0), s(_x1))]] = 3 >= 0 = [[s(_x1)]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 2 + x2 + x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 3F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 2 + x1 + x2 + 2F0(x2) + x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + x1 + x2 + 2F0(x2) + x2F0(x2) >= 2 + x1 + x2 + F0(x2) + x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + x1 + x2 + 2F0(x2) + x2F0(x2) >= x2 + F0(x2) + x2F0(x2) = [[filter(_F0, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_2, R_1, minimal, formative), where P_2 consists of:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_1, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  
    * 1 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.