We consider the system zipWith.

  Alphabet:

    0 : [] --> nat 
    cons : [nat * list] --> list 
    false : [] --> bool 
    gcd : [nat * nat] --> nat 
    gcdlists : [list * list] --> list 
    if : [bool * nat * nat] --> nat 
    le : [nat * nat] --> bool 
    minus : [nat * nat] --> nat 
    nil : [] --> list 
    s : [nat] --> nat 
    true : [] --> bool 
    zipWith : [nat -> nat -> nat * list * list] --> list 

  Rules:

    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    minus(x, 0) => x 
    minus(s(x), s(y)) => minus(x, y) 
    gcd(0, x) => 0 
    gcd(s(x), 0) => 0 
    gcd(s(x), s(y)) => if(le(y, x), s(x), s(y)) 
    if(true, s(x), s(y)) => gcd(minus(x, y), s(y)) 
    if(false, s(x), s(y)) => gcd(minus(y, x), s(x)) 
    zipWith(f, x, nil) => nil 
    zipWith(f, nil, x) => nil 
    zipWith(f, cons(x, y), cons(z, u)) => cons(f x z, zipWith(f, y, u)) 
    gcdlists(x, y) => zipWith(/\z./\u.gcd(z, u), x, 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:

  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 
  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  gcd(0, X) => 0 
  gcd(s(X), 0) => 0 
  gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) 
  if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) 
  if(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) 

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
 ||     (15) QDP
 ||         (16) QDPOrderProof [EQUIVALENT]
 ||         (17) QDP
 ||         (18) PisEmptyProof [EQUIVALENT]
 ||         (19) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~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:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 ||    GCD(s(%X), s(%Y)) -> IF(le(%Y, %X), s(%X), s(%Y))
 ||    GCD(s(%X), s(%Y)) -> LE(%Y, %X)
 ||    IF(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y))
 ||    IF(true, s(%X), s(%Y)) -> MINUS(%X, %Y)
 ||    IF(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X))
 ||    IF(false, s(%X), s(%Y)) -> MINUS(%Y, %X)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~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 3 SCCs with 3 less nodes.
 || ----------------------------------------
 || 
 || (4)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (5)
 || 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:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~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:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || 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:
 || *MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (9)
 || YES
 || 
 || ----------------------------------------
 || 
 || (10)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~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:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%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:
 || *LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (14)
 || YES
 || 
 || ----------------------------------------
 || 
 || (15)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    GCD(s(%X), s(%Y)) -> IF(le(%Y, %X), s(%X), s(%Y))
 ||    IF(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y))
 ||    IF(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (16) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    GCD(s(%X), s(%Y)) -> IF(le(%Y, %X), s(%X), s(%Y))
 ||    IF(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y))
 ||    IF(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
 || 
 || POL( IF_3(x_1, ..., x_3) ) = 2x_2 + x_3 + 1
 || POL( le_2(x_1, x_2) ) = 0
 || POL( 0 ) = 0
 || POL( true ) = 2
 || POL( s_1(x_1) ) = 2x_1 + 2
 || POL( false ) = 1
 || POL( GCD_2(x_1, x_2) ) = 2x_1 + x_2 + 2
 || POL( minus_2(x_1, x_2) ) = x_1
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 || 
 || 
 || ----------------------------------------
 || 
 || (17)
 || Obligation:
 || Q DP problem:
 || P is empty.
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    gcd(0, %X) -> 0
 ||    gcd(s(%X), 0) -> 0
 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (18) PisEmptyProof (EQUIVALENT)
 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
 || ----------------------------------------
 || 
 || (19)
 || 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] zipWith#(F, cons(X, Y), cons(Z, U)) =#> F(X, Z)   
    1] zipWith#(F, cons(X, Y), cons(Z, U)) =#> zipWith#(F, Y, U)   
    2] gcdlists#(X, Y) =#> zipWith#(/\x./\y.gcd(x, y), X, Y)   
    3] gcdlists#(X, Y) =#> gcd#(x, y)   

  Rules R_0:

    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    gcd(0, X) => 0 
    gcd(s(X), 0) => 0 
    gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) 
    if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) 
    if(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) 
    zipWith(F, X, nil) => nil 
    zipWith(F, nil, X) => nil 
    zipWith(F, cons(X, Y), cons(Z, U)) => cons(F X Z, zipWith(F, Y, U)) 
    gcdlists(X, Y) => zipWith(/\x./\y.gcd(x, y), X, 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).

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, 3 
    * 1 : 0, 1 
    * 2 : 0, 1 
    * 3 :  

This graph has the following strongly connected components:

  P_1:

    zipWith#(F, cons(X, Y), cons(Z, U)) =#> F(X, Z)   
    zipWith#(F, cons(X, Y), cons(Z, U)) =#> zipWith#(F, Y, U)   
    gcdlists#(X, Y) =#> zipWith#(/\x./\y.gcd(x, y), X, Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f).

Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

The formative rules of (P_1, R_0) are R_1 ::=

  zipWith(F, cons(X, Y), cons(Z, U)) => cons(F X Z, zipWith(F, Y, U)) 
  gcdlists(X, Y) => zipWith(/\x./\y.gcd(x, y), X, Y) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative).

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:

  zipWith#(F, cons(X, Y), cons(Z, U)) >? F(X, Z) 
  zipWith#(F, cons(X, Y), cons(Z, U)) >? zipWith#(F, Y, U) 
  gcdlists#(X, Y) >? zipWith#(/\x./\y.gcd-(x, y), X, Y) 
  zipWith(F, cons(X, Y), cons(Z, U)) >= cons(F X Z, zipWith(F, Y, U)) 
  gcdlists(X, Y) >= zipWith(/\x./\y.gcd-(x, y), X, Y) 
  gcd-(X, Y) >= gcd(X, Y) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( gcdlists(X, Y) ) = #argfun-gcdlists#(zipWith(/\x./\y.gcd-(x, y), X, Y)) 
  pi( gcdlists#(X, Y) ) = #argfun-gcdlists##(zipWith#(/\x./\y.gcd-(x, y), X, Y)) 

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

The following interpretation satisfies the requirements:

  #argfun-gcdlists# = \y0.3 + y0 
  #argfun-gcdlists## = \y0.3 + y0 
  cons = \y0y1.1 + y0 + y1 
  gcd = \y0y1.0 
  gcd- = \y0y1.3 + 3y0 + 3y1 
  gcdlists = \y0y1.0 
  gcdlists# = \y0y1.0 
  zipWith = \G0y1y2.y1 + 2y2 + 2y1y2G0(y1,y2) + 2G0(y1,y2) 
  zipWith# = \G0y1y2.3 + G0(y1,y2) 

Using this interpretation, the requirements translate to:

  [[zipWith#(_F0, cons(_x1, _x2), cons(_x3, _x4))]] = 3 + F0(1 + x1 + x2,1 + x3 + x4) > F0(x1,x3) = [[_F0(_x1, _x3)]] 
  [[zipWith#(_F0, cons(_x1, _x2), cons(_x3, _x4))]] = 3 + F0(1 + x1 + x2,1 + x3 + x4) >= 3 + F0(x2,x4) = [[zipWith#(_F0, _x2, _x4)]] 
  [[#argfun-gcdlists##(zipWith#(/\x./\y.gcd-(x, y), _x0, _x1))]] = 9 + 3x0 + 3x1 > 6 + 3x0 + 3x1 = [[zipWith#(/\x./\y.gcd-(x, y), _x0, _x1)]] 
  [[zipWith(_F0, cons(_x1, _x2), cons(_x3, _x4))]] = 3 + x1 + x2 + 2x3 + 2x4 + 2x1x3F0(1 + x1 + x2,1 + x3 + x4) + 2x1x4F0(1 + x1 + x2,1 + x3 + x4) + 2x1F0(1 + x1 + x2,1 + x3 + x4) + 2x2x3F0(1 + x1 + x2,1 + x3 + x4) + 2x2x4F0(1 + x1 + x2,1 + x3 + x4) + 2x2F0(1 + x1 + x2,1 + x3 + x4) + 2x3F0(1 + x1 + x2,1 + x3 + x4) + 2x4F0(1 + x1 + x2,1 + x3 + x4) + 4F0(1 + x1 + x2,1 + x3 + x4) >= 1 + x2 + 2x4 + 2x2x4F0(x2,x4) + 2F0(x2,x4) + max(x1, x3, F0(x1,x3)) = [[cons(_F0 _x1 _x3, zipWith(_F0, _x2, _x4))]] 
  [[#argfun-gcdlists#(zipWith(/\x./\y.gcd-(x, y), _x0, _x1))]] = 9 + 6x0x0x1 + 6x0x1 + 6x0x1x1 + 7x0 + 8x1 >= 6 + 6x0x0x1 + 6x0x1 + 6x0x1x1 + 7x0 + 8x1 = [[zipWith(/\x./\y.gcd-(x, y), _x0, _x1)]] 
  [[gcd-(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 0 = [[gcd(_x0, _x1)]] 

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:

  zipWith#(F, cons(X, Y), cons(Z, U)) =#> zipWith#(F, Y, U)   

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(zipWith#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(zipWith#(F, cons(X, Y), cons(Z, U))) = cons(X, Y) |> Y = nu(zipWith#(F, Y, U)) 

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.