We consider the system Applicative_first_order_05__perfect.

  Alphabet:

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

  Rules:

    perfectp(0) => false 
    perfectp(s(x)) => f(x, s(0), s(x), s(x)) 
    f(0, x, 0, y) => true 
    f(0, x, s(y), z) => false 
    f(s(x), 0, y, z) => f(x, z, minus(y, s(x)), z) 
    f(s(x), s(y), z, u) => if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) 
    map(g, nil) => nil 
    map(g, cons(x, y)) => cons(g x, map(g, y)) 
    filter(g, nil) => nil 
    filter(g, cons(x, y)) => filter2(g x, g, x, y) 
    filter2(true, g, x, y) => cons(x, filter(g, y)) 
    filter2(false, g, x, y) => filter(g, 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:

  perfectp(0) => false 
  perfectp(s(X)) => f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) => true 
  f(0, X, s(Y), Z) => false 
  f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 

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) QDP
 || (7) UsableRulesProof [EQUIVALENT]
 || (8) QDP
 || (9) QReductionProof [EQUIVALENT]
 || (10) QDP
 || (11) QDPSizeChangeProof [EQUIVALENT]
 || (12) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    perfectp(0) -> false
 ||    perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X))
 ||    f(0, %X, 0, %Y) -> true
 ||    f(0, %X, s(%Y), %Z) -> false
 ||    f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 || 
 || 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:
 || 
 ||    perfectp(0) -> false
 ||    perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X))
 ||    f(0, %X, 0, %Y) -> true
 ||    f(0, %X, s(%Y), %Z) -> false
 ||    f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 || 
 || The set Q consists of the following terms:
 || 
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 
 || ----------------------------------------
 || 
 || (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:
 || 
 ||    PERFECTP(s(%X)) -> F(%X, s(0), s(%X), s(%X))
 ||    F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    F(s(%X), s(%Y), %Z, %U) -> F(s(%X), minus(%Y, %X), %Z, %U)
 ||    F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    perfectp(0) -> false
 ||    perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X))
 ||    f(0, %X, 0, %Y) -> true
 ||    f(0, %X, s(%Y), %Z) -> false
 ||    f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 || 
 || The set Q consists of the following terms:
 || 
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (5) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
 || ----------------------------------------
 || 
 || (6)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U)
 ||    F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    perfectp(0) -> false
 ||    perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X))
 ||    f(0, %X, 0, %Y) -> true
 ||    f(0, %X, s(%Y), %Z) -> false
 ||    f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 || 
 || The set Q consists of the following terms:
 || 
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (7) 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.
 || ----------------------------------------
 || 
 || (8)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U)
 ||    F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (9) 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].
 || 
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 
 || ----------------------------------------
 || 
 || (10)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U)
 ||    F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (11) 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), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U)
 || The graph contains the following edges 1 > 1, 4 >= 2, 3 >= 3, 4 >= 4
 || 
 || 
 || *F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z)
 || The graph contains the following edges 1 > 1, 4 >= 2, 4 >= 4
 || 
 || 
 || ----------------------------------------
 || 
 || (12)
 || 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:

    perfectp(0) => false 
    perfectp(s(X)) => f(X, s(0), s(X), s(X)) 
    f(0, X, 0, Y) => true 
    f(0, X, s(Y), Z) => false 
    f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 
    f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
    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 ::=

  perfectp(0) => false 
  perfectp(s(X)) => f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) => true 
  f(0, X, s(Y), Z) => false 
  f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 
  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) 
  perfectp(0) >= false 
  perfectp(s(X)) >= f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) >= true 
  f(0, X, s(Y), Z) >= false 
  f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 
  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 
  cons = \y0y1.2 + 2y0 + 2y1 
  f = \y0y1y2y3.0 
  false = 0 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.2 + 2y2 + 2y3 
  filter2# = \y0G1y2y3.2y3G1(y3) 
  filter# = \G0y1.2y1G0(y1) 
  map = \G0y1.y1 + 2y1G0(y1) 
  map# = \G0y1.3 + 2G0(y1) 
  minus = \y0y1.0 
  perfectp = \y0.3 
  s = \y0.0 
  true = 0 

Using this interpretation, the requirements translate to:

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

This graph has the following strongly connected components:

  P_2:

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

  P_3:

    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)   

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) and (P_3, R_1, m, f).

Thus, the original system is terminating if each of (P_2, R_1, minimal, formative) and (P_3, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_3, 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) 
  perfectp(0) >= false 
  perfectp(s(X)) >= f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) >= true 
  f(0, X, s(Y), Z) >= false 
  f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 
  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 
  cons = \y0y1.2 + y0 + y1 
  f = \y0y1y2y3.2 
  false = 0 
  filter = \G0y1.2y1 
  filter2 = \y0G1y2y3.2 + y2 + 2y3 
  filter2# = \y0G1y2y3.2y3 + y3G1(y3) 
  filter# = \G0y1.2y1 + y1G0(y1) 
  map = \G0y1.y1 + 2y1G0(y1) 
  minus = \y0y1.0 
  perfectp = \y0.3 
  s = \y0.0 
  true = 0 

Using this interpretation, the requirements translate to:

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

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_1, minimal, formative) by (P_4, R_1, minimal, formative), where P_4 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 each of (P_2, R_1, minimal, formative) and (P_4, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_4, 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.

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.