We consider the system Applicative_first_order_05__#3.36.

  Alphabet:

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

  Rules:

    minus(x, 0) => x 
    minus(s(x), s(y)) => minus(x, y) 
    f(0) => s(0) 
    f(s(x)) => minus(s(x), g(f(x))) 
    g(0) => 0 
    g(s(x)) => minus(s(x), f(g(x))) 
    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:

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  f(0) => s(0) 
  f(s(X)) => minus(s(X), g(f(X))) 
  g(0) => 0 
  g(s(X)) => minus(s(X), f(g(X))) 

Moreover, the system is orthogonal.  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 terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed 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) Overlay + Local Confluence [EQUIVALENT]
 || (2) QTRS
 || (3) DependencyPairsProof [EQUIVALENT]
 || (4) QDP
 || (5) DependencyGraphProof [EQUIVALENT]
 || (6) AND
 ||     (7) QDP
 ||         (8) UsableRulesProof [EQUIVALENT]
 ||         (9) QDP
 ||         (10) QReductionProof [EQUIVALENT]
 ||         (11) QDP
 ||         (12) QDPSizeChangeProof [EQUIVALENT]
 ||         (13) YES
 ||     (14) QDP
 ||         (15) QDPOrderProof [EQUIVALENT]
 ||         (16) QDP
 ||         (17) DependencyGraphProof [EQUIVALENT]
 ||         (18) TRUE
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    f(0) -> s(0)
 ||    f(s(%X)) -> minus(s(%X), g(f(%X)))
 ||    g(0) -> 0
 ||    g(s(%X)) -> minus(s(%X), f(g(%X)))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) Overlay + Local Confluence (EQUIVALENT)
 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    f(0) -> s(0)
 ||    f(s(%X)) -> minus(s(%X), g(f(%X)))
 ||    g(0) -> 0
 ||    g(s(%X)) -> minus(s(%X), f(g(%X)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    f(0)
 ||    f(s(x0))
 ||    g(0)
 ||    g(s(x0))
 || 
 || 
 || ----------------------------------------
 || 
 || (3) DependencyPairsProof (EQUIVALENT)
 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 ||    F(s(%X)) -> MINUS(s(%X), g(f(%X)))
 ||    F(s(%X)) -> G(f(%X))
 ||    F(s(%X)) -> F(%X)
 ||    G(s(%X)) -> MINUS(s(%X), f(g(%X)))
 ||    G(s(%X)) -> F(g(%X))
 ||    G(s(%X)) -> G(%X)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    f(0) -> s(0)
 ||    f(s(%X)) -> minus(s(%X), g(f(%X)))
 ||    g(0) -> 0
 ||    g(s(%X)) -> minus(s(%X), f(g(%X)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    f(0)
 ||    f(s(x0))
 ||    g(0)
 ||    g(s(x0))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (5) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.
 || ----------------------------------------
 || 
 || (6)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (7)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    f(0) -> s(0)
 ||    f(s(%X)) -> minus(s(%X), g(f(%X)))
 ||    g(0) -> 0
 ||    g(s(%X)) -> minus(s(%X), f(g(%X)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    f(0)
 ||    f(s(x0))
 ||    g(0)
 ||    g(s(x0))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (8) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (9)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    f(0)
 ||    f(s(x0))
 ||    g(0)
 ||    g(s(x0))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (10) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    f(0)
 ||    f(s(x0))
 ||    g(0)
 ||    g(s(x0))
 || 
 || 
 || ----------------------------------------
 || 
 || (11)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (12) 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:
 || *MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (13)
 || YES
 || 
 || ----------------------------------------
 || 
 || (14)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    F(s(%X)) -> G(f(%X))
 ||    G(s(%X)) -> F(g(%X))
 ||    F(s(%X)) -> F(%X)
 ||    G(s(%X)) -> G(%X)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    f(0) -> s(0)
 ||    f(s(%X)) -> minus(s(%X), g(f(%X)))
 ||    g(0) -> 0
 ||    g(s(%X)) -> minus(s(%X), f(g(%X)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    f(0)
 ||    f(s(x0))
 ||    g(0)
 ||    g(s(x0))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (15) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    G(s(%X)) -> F(g(%X))
 ||    F(s(%X)) -> F(%X)
 ||    G(s(%X)) -> G(%X)
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(F(x_1)) = x_1
 ||    POL(G(x_1)) = x_1
 ||    POL(f(x_1)) = 1 + x_1
 ||    POL(g(x_1)) = x_1
 ||    POL(minus(x_1, x_2)) = x_1
 ||    POL(s(x_1)) = 1 + x_1
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    f(0) -> s(0)
 ||    f(s(%X)) -> minus(s(%X), g(f(%X)))
 ||    g(0) -> 0
 ||    g(s(%X)) -> minus(s(%X), f(g(%X)))
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 || 
 || 
 || ----------------------------------------
 || 
 || (16)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    F(s(%X)) -> G(f(%X))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    f(0) -> s(0)
 ||    f(s(%X)) -> minus(s(%X), g(f(%X)))
 ||    g(0) -> 0
 ||    g(s(%X)) -> minus(s(%X), f(g(%X)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    f(0)
 ||    f(s(x0))
 ||    g(0)
 ||    g(s(x0))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (17) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
 || ----------------------------------------
 || 
 || (18)
 || TRUE
 || 
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:

    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    f(0) => s(0) 
    f(s(X)) => minus(s(X), g(f(X))) 
    g(0) => 0 
    g(s(X)) => minus(s(X), f(g(X))) 
    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 ::=

  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) 
  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:

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

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) >= 3 + 2F0(x2) + x2F0(x2) = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 6F0(2 + x2 + 2x1) > x2 + 2x2F0(x2) + 2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 6F0(2 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = x2 + 2x2F0(x2) + 2F0(x2) >= x2 + 2x2F0(x2) + 2F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = x2 + 2x2F0(x2) + 2F0(x2) >= x2 + 2x2F0(x2) + 2F0(x2) = [[filter#(_F0, _x2)]] 
  [[map(_F0, cons(_x1, _x2))]] = 4 + 2x2 + 4x1 + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) >= 2 + 2x2 + x2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) >= 2 + x2 + 2x1 + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + x2 + 2x1 + 2x2F0(x2) >= 2 + x2 + 2x1 + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + x2 + 2x1 + 2x2F0(x2) >= x2 + 2x2F0(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:

  map#(F, cons(X, Y)) =#> map#(F, Y)   
  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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 0 
    * 1 :  
    * 2 :  

This graph has the following strongly connected components:

  P_2:

    map#(F, cons(X, Y)) =#> map#(F, Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_1, R_1, m, f) by (P_2, R_1, m, f).

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 apply the subterm criterion with the following projection function:

  nu(map#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_1, minimal, f) by ({}, R_1, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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


+++ Citations +++

[FuhKop19]  C. Fuhs, and C. Kop.  A static higher-order dependency pair framework.  In Proceedings of ESOP 2019, 2019.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.