We consider the system Applicative_first_order_05__12.

  Alphabet:

    and : [a * a] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    not : [a] --> a 
    or : [a * a] --> a 
    true : [] --> b 

  Rules:

    not(not(x)) => x 
    not(or(x, y)) => and(not(x), not(y)) 
    not(and(x, y)) => or(not(x), not(y)) 
    and(x, or(y, z)) => or(and(x, y), and(x, z)) 
    and(or(x, y), z) => or(and(z, x), and(z, 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) 

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

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).

We thus obtain the following dependency pair problem (P_0, R_0, static, formative):

  Dependency Pairs P_0:

    0] not#(or(X, Y)) =#> and#(not(X), not(Y))   
    1] not#(or(X, Y)) =#> not#(X)   
    2] not#(or(X, Y)) =#> not#(Y)   
    3] not#(and(X, Y)) =#> not#(X)   
    4] not#(and(X, Y)) =#> not#(Y)   
    5] and#(X, or(Y, Z)) =#> and#(X, Y)   
    6] and#(X, or(Y, Z)) =#> and#(X, Z)   
    7] and#(or(X, Y), Z) =#> and#(Z, X)   
    8] and#(or(X, Y), Z) =#> and#(Z, Y)   
    9] map#(F, cons(X, Y)) =#> map#(F, Y)   
    10] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    11] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    12] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    not(not(X)) => X 
    not(or(X, Y)) => and(not(X), not(Y)) 
    not(and(X, Y)) => or(not(X), not(Y)) 
    and(X, or(Y, Z)) => or(and(X, Y), and(X, Z)) 
    and(or(X, Y), Z) => or(and(Z, X), and(Z, 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, static, formative) is finite.

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

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

  not(not(X)) => X 
  not(or(X, Y)) => and(not(X), not(Y)) 
  not(and(X, Y)) => or(not(X), not(Y)) 
  and(X, or(Y, Z)) => or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) => or(and(Z, X), and(Z, 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, static, formative) by (P_0, R_1, static, formative).

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

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  not#(or(X, Y)) >? and#(not(X), not(Y)) 
  not#(or(X, Y)) >? not#(X) 
  not#(or(X, Y)) >? not#(Y) 
  not#(and(X, Y)) >? not#(X) 
  not#(and(X, Y)) >? not#(Y) 
  and#(X, or(Y, Z)) >? and#(X, Y) 
  and#(X, or(Y, Z)) >? and#(X, Z) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  and#(or(X, Y), Z) >? and#(Z, Y) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  filter2#(false, F, X, Y) >? filter#(F, Y) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, 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:

  and = \y0y1.0 
  and# = \y0y1.0 
  cons = \y0y1.0 
  false = 3 
  filter = \G0y1.0 
  filter2 = \y0G1y2y3.0 
  filter2# = \y0G1y2y3.0 
  filter# = \G0y1.0 
  map = \G0y1.2G0(0) 
  map# = \G0y1.0 
  not = \y0.y0 
  not# = \y0.3 
  or = \y0y1.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[not#(or(_x0, _x1))]] = 3 > 0 = [[and#(not(_x0), not(_x1))]] 
  [[not#(or(_x0, _x1))]] = 3 >= 3 = [[not#(_x0)]] 
  [[not#(or(_x0, _x1))]] = 3 >= 3 = [[not#(_x1)]] 
  [[not#(and(_x0, _x1))]] = 3 >= 3 = [[not#(_x0)]] 
  [[not#(and(_x0, _x1))]] = 3 >= 3 = [[not#(_x1)]] 
  [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x1)]] 
  [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x2)]] 
  [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x0)]] 
  [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] 
  [[not(not(_x0))]] = x0 >= x0 = [[_x0]] 
  [[not(or(_x0, _x1))]] = 0 >= 0 = [[and(not(_x0), not(_x1))]] 
  [[not(and(_x0, _x1))]] = 0 >= 0 = [[or(not(_x0), not(_x1))]] 
  [[and(_x0, or(_x1, _x2))]] = 0 >= 0 = [[or(and(_x0, _x1), and(_x0, _x2))]] 
  [[and(or(_x0, _x1), _x2)]] = 0 >= 0 = [[or(and(_x2, _x0), and(_x2, _x1))]] 
  [[map(_F0, cons(_x1, _x2))]] = 2F0(0) >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 0 >= 0 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 0 >= 0 = [[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, static, formative) by (P_1, R_1, static, formative), where P_1 consists of:

  not#(or(X, Y)) =#> not#(X)   
  not#(or(X, Y)) =#> not#(Y)   
  not#(and(X, Y)) =#> not#(X)   
  not#(and(X, Y)) =#> not#(Y)   
  and#(X, or(Y, Z)) =#> and#(X, Y)   
  and#(X, or(Y, Z)) =#> and#(X, Z)   
  and#(or(X, Y), Z) =#> and#(Z, X)   
  and#(or(X, Y), Z) =#> and#(Z, Y)   
  map#(F, cons(X, Y)) =#> map#(F, Y)   
  filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, 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, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  not#(or(X, Y)) >? not#(X) 
  not#(or(X, Y)) >? not#(Y) 
  not#(and(X, Y)) >? not#(X) 
  not#(and(X, Y)) >? not#(Y) 
  and#(X, or(Y, Z)) >? and#(X, Y) 
  and#(X, or(Y, Z)) >? and#(X, Z) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  and#(or(X, Y), Z) >? and#(Z, Y) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  filter2#(false, F, X, Y) >? filter#(F, Y) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, 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:

  and = \y0y1.0 
  and# = \y0y1.0 
  cons = \y0y1.2 + 2y1 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.2 + 2y3 
  filter2# = \y0G1y2y3.2y3G1(y3) + 2G1(0) 
  filter# = \G0y1.2y1G0(y1) 
  map = \G0y1.y1 
  map# = \G0y1.2y1 
  not = \y0.y0 
  not# = \y0.0 
  or = \y0y1.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[not#(or(_x0, _x1))]] = 0 >= 0 = [[not#(_x0)]] 
  [[not#(or(_x0, _x1))]] = 0 >= 0 = [[not#(_x1)]] 
  [[not#(and(_x0, _x1))]] = 0 >= 0 = [[not#(_x0)]] 
  [[not#(and(_x0, _x1))]] = 0 >= 0 = [[not#(_x1)]] 
  [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x1)]] 
  [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x2)]] 
  [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x0)]] 
  [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 4 + 4x2 > 2x2 = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 4x2F0(2 + 2x2) + 4F0(2 + 2x2) >= 2x2F0(x2) + 2F0(0) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 2x2F0(x2) + 2F0(0) >= 2x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 2x2F0(x2) + 2F0(0) >= 2x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[not(not(_x0))]] = x0 >= x0 = [[_x0]] 
  [[not(or(_x0, _x1))]] = 0 >= 0 = [[and(not(_x0), not(_x1))]] 
  [[not(and(_x0, _x1))]] = 0 >= 0 = [[or(not(_x0), not(_x1))]] 
  [[and(_x0, or(_x1, _x2))]] = 0 >= 0 = [[or(and(_x0, _x1), and(_x0, _x2))]] 
  [[and(or(_x0, _x1), _x2)]] = 0 >= 0 = [[or(and(_x2, _x0), and(_x2, _x1))]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 2 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 2 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + 2x2 >= 2 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + 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_1, R_1, static, formative) by (P_2, R_1, static, formative), where P_2 consists of:

  not#(or(X, Y)) =#> not#(X)   
  not#(or(X, Y)) =#> not#(Y)   
  not#(and(X, Y)) =#> not#(X)   
  not#(and(X, Y)) =#> not#(Y)   
  and#(X, or(Y, Z)) =#> and#(X, Y)   
  and#(X, or(Y, Z)) =#> and#(X, Z)   
  and#(or(X, Y), Z) =#> and#(Z, X)   
  and#(or(X, Y), Z) =#> and#(Z, Y)   
  filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, 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_2, R_1, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  not#(or(X, Y)) >? not#(X) 
  not#(or(X, Y)) >? not#(Y) 
  not#(and(X, Y)) >? not#(X) 
  not#(and(X, Y)) >? not#(Y) 
  and#(X, or(Y, Z)) >? and#(X, Y) 
  and#(X, or(Y, Z)) >? and#(X, Z) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  and#(or(X, Y), Z) >? and#(Z, Y) 
  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  filter2#(false, F, X, Y) >? filter#(F, Y) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, 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:

  and = \y0y1.0 
  and# = \y0y1.0 
  cons = \y0y1.1 + y1 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.1 + y3 
  filter2# = \y0G1y2y3.1 + 2y3 
  filter# = \G0y1.2y1 
  map = \G0y1.y1 
  not = \y0.2y0 
  not# = \y0.0 
  or = \y0y1.0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[not#(or(_x0, _x1))]] = 0 >= 0 = [[not#(_x0)]] 
  [[not#(or(_x0, _x1))]] = 0 >= 0 = [[not#(_x1)]] 
  [[not#(and(_x0, _x1))]] = 0 >= 0 = [[not#(_x0)]] 
  [[not#(and(_x0, _x1))]] = 0 >= 0 = [[not#(_x1)]] 
  [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x1)]] 
  [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x2)]] 
  [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x0)]] 
  [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x1)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 2 + 2x2 > 1 + 2x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 1 + 2x2 > 2x2 = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 1 + 2x2 > 2x2 = [[filter#(_F0, _x2)]] 
  [[not(not(_x0))]] = 4x0 >= x0 = [[_x0]] 
  [[not(or(_x0, _x1))]] = 0 >= 0 = [[and(not(_x0), not(_x1))]] 
  [[not(and(_x0, _x1))]] = 0 >= 0 = [[or(not(_x0), not(_x1))]] 
  [[and(_x0, or(_x1, _x2))]] = 0 >= 0 = [[or(and(_x0, _x1), and(_x0, _x2))]] 
  [[and(or(_x0, _x1), _x2)]] = 0 >= 0 = [[or(and(_x2, _x0), and(_x2, _x1))]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 1 + x2 >= 1 + x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 1 + 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_2, R_1, static, formative) by (P_3, R_1, static, formative), where P_3 consists of:

  not#(or(X, Y)) =#> not#(X)   
  not#(or(X, Y)) =#> not#(Y)   
  not#(and(X, Y)) =#> not#(X)   
  not#(and(X, Y)) =#> not#(Y)   
  and#(X, or(Y, Z)) =#> and#(X, Y)   
  and#(X, or(Y, Z)) =#> and#(X, Z)   
  and#(or(X, Y), Z) =#> and#(Z, X)   
  and#(or(X, Y), Z) =#> and#(Z, Y)   

Thus, the original system is terminating if (P_3, R_1, static, formative) is finite.

We consider the dependency pair problem (P_3, R_1, static, formative).

The formative rules of (P_3, R_1) are R_2 ::=

  not(not(X)) => X 
  not(or(X, Y)) => and(not(X), not(Y)) 
  not(and(X, Y)) => or(not(X), not(Y)) 
  and(X, or(Y, Z)) => or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) => or(and(Z, X), and(Z, Y)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_3, R_1, static, formative) by (P_3, R_2, static, formative).

Thus, the original system is terminating if (P_3, R_2, static, formative) is finite.

We consider the dependency pair problem (P_3, R_2, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  not#(or(X, Y)) >? not#(X) 
  not#(or(X, Y)) >? not#(Y) 
  not#(and(X, Y)) >? not#(X) 
  not#(and(X, Y)) >? not#(Y) 
  and#(X, or(Y, Z)) >? and#(X, Y) 
  and#(X, or(Y, Z)) >? and#(X, Z) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  and#(or(X, Y), Z) >? and#(Z, Y) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {and, and#, not, not#, or}, and the following precedence: and# > not > not# > and > or

With these choices, we have:

  1] not#(or(X, Y)) >= not#(X)  because not# in Mul and [2], by (Fun) 
  2] or(X, Y) >= X  because [3], by (Star) 
  3] or*(X, Y) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

  5] not#(or(X, Y)) >= not#(Y)  because not# in Mul and [6], by (Fun) 
  6] or(X, Y) >= Y  because [7], by (Star) 
  7] or*(X, Y) >= Y  because [8], by (Select) 
  8] Y >= Y  by (Meta) 

  9] not#(and(X, Y)) >= not#(X)  because not# in Mul and [10], by (Fun) 
  10] and(X, Y) >= X  because [11], by (Star) 
  11] and*(X, Y) >= X  because [12], by (Select) 
  12] X >= X  by (Meta) 

  13] not#(and(X, Y)) >= not#(Y)  because not# in Mul and [14], by (Fun) 
  14] and(X, Y) >= Y  because [15], by (Star) 
  15] and*(X, Y) >= Y  because [16], by (Select) 
  16] Y >= Y  by (Meta) 

  17] and#(X, or(Y, Z)) >= and#(X, Y)  because and# in Mul, [18] and [19], by (Fun) 
  18] X >= X  by (Meta) 
  19] or(Y, Z) >= Y  because [20], by (Star) 
  20] or*(Y, Z) >= Y  because [21], by (Select) 
  21] Y >= Y  by (Meta) 

  22] and#(X, or(Y, Z)) >= and#(X, Z)  because and# in Mul, [18] and [23], by (Fun) 
  23] or(Y, Z) >= Z  because [24], by (Star) 
  24] or*(Y, Z) >= Z  because [25], by (Select) 
  25] Z >= Z  by (Meta) 

  26] and#(or(X, Y), Z) >= and#(Z, X)  because and# in Mul, [27] and [30], by (Fun) 
  27] or(X, Y) >= X  because [28], by (Star) 
  28] or*(X, Y) >= X  because [29], by (Select) 
  29] X >= X  by (Meta) 
  30] Z >= Z  by (Meta) 

  31] and#(or(X, Y), Z) > and#(Z, Y)  because [32], by definition 
  32] and#*(or(X, Y), Z) >= and#(Z, Y)  because and# in Mul, [33] and [30], by (Stat) 
  33] or(X, Y) > Y  because [34], by definition 
  34] or*(X, Y) >= Y  because [35], by (Select) 
  35] Y >= Y  by (Meta) 

  36] not(not(X)) >= X  because [37], by (Star) 
  37] not*(not(X)) >= X  because [38], by (Select) 
  38] not(X) >= X  because [39], by (Star) 
  39] not*(X) >= X  because [40], by (Select) 
  40] X >= X  by (Meta) 

  41] not(or(X, Y)) >= and(not(X), not(Y))  because [42], by (Star) 
  42] not*(or(X, Y)) >= and(not(X), not(Y))  because not > and, [43] and [46], by (Copy) 
  43] not*(or(X, Y)) >= not(X)  because not in Mul and [44], by (Stat) 
  44] or(X, Y) > X  because [45], by definition 
  45] or*(X, Y) >= X  because [4], by (Select) 
  46] not*(or(X, Y)) >= not(Y)  because not in Mul and [47], by (Stat) 
  47] or(X, Y) > Y  because [48], by definition 
  48] or*(X, Y) >= Y  because [8], by (Select) 

  49] not(and(X, Y)) >= or(not(X), not(Y))  because [50], by (Star) 
  50] not*(and(X, Y)) >= or(not(X), not(Y))  because not > or, [51] and [54], by (Copy) 
  51] not*(and(X, Y)) >= not(X)  because not in Mul and [52], by (Stat) 
  52] and(X, Y) > X  because [53], by definition 
  53] and*(X, Y) >= X  because [12], by (Select) 
  54] not*(and(X, Y)) >= not(Y)  because not in Mul and [55], by (Stat) 
  55] and(X, Y) > Y  because [56], by definition 
  56] and*(X, Y) >= Y  because [16], by (Select) 

  57] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because [58], by (Star) 
  58] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because and > or, [59] and [62], by (Copy) 
  59] and*(X, or(Y, Z)) >= and(X, Y)  because and in Mul, [18] and [60], by (Stat) 
  60] or(Y, Z) > Y  because [61], by definition 
  61] or*(Y, Z) >= Y  because [21], by (Select) 
  62] and*(X, or(Y, Z)) >= and(X, Z)  because and in Mul, [18] and [63], by (Stat) 
  63] or(Y, Z) > Z  because [64], by definition 
  64] or*(Y, Z) >= Z  because [25], by (Select) 

  65] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because [66], by (Star) 
  66] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because and > or, [67] and [70], by (Copy) 
  67] and*(or(X, Y), Z) >= and(Z, X)  because and in Mul, [68] and [30], by (Stat) 
  68] or(X, Y) > X  because [69], by definition 
  69] or*(X, Y) >= X  because [29], by (Select) 
  70] and*(or(X, Y), Z) >= and(Z, Y)  because and in Mul, [33] and [30], by (Stat) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_2, static, formative) by (P_4, R_2, static, formative), where P_4 consists of:

  not#(or(X, Y)) =#> not#(X)   
  not#(or(X, Y)) =#> not#(Y)   
  not#(and(X, Y)) =#> not#(X)   
  not#(and(X, Y)) =#> not#(Y)   
  and#(X, or(Y, Z)) =#> and#(X, Y)   
  and#(X, or(Y, Z)) =#> and#(X, Z)   
  and#(or(X, Y), Z) =#> and#(Z, X)   

Thus, the original system is terminating if (P_4, R_2, static, formative) is finite.

We consider the dependency pair problem (P_4, R_2, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  not#(or(X, Y)) >? not#(X) 
  not#(or(X, Y)) >? not#(Y) 
  not#(and(X, Y)) >? not#(X) 
  not#(and(X, Y)) >? not#(Y) 
  and#(X, or(Y, Z)) >? and#(X, Y) 
  and#(X, or(Y, Z)) >? and#(X, Z) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {and, and#, not, not#, or}, and the following precedence: and# > not > not# > and > or

With these choices, we have:

  1] not#(or(X, Y)) >= not#(X)  because not# in Mul and [2], by (Fun) 
  2] or(X, Y) >= X  because [3], by (Star) 
  3] or*(X, Y) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

  5] not#(or(X, Y)) >= not#(Y)  because not# in Mul and [6], by (Fun) 
  6] or(X, Y) >= Y  because [7], by (Star) 
  7] or*(X, Y) >= Y  because [8], by (Select) 
  8] Y >= Y  by (Meta) 

  9] not#(and(X, Y)) >= not#(X)  because not# in Mul and [10], by (Fun) 
  10] and(X, Y) >= X  because [11], by (Star) 
  11] and*(X, Y) >= X  because [12], by (Select) 
  12] X >= X  by (Meta) 

  13] not#(and(X, Y)) >= not#(Y)  because not# in Mul and [14], by (Fun) 
  14] and(X, Y) >= Y  because [15], by (Star) 
  15] and*(X, Y) >= Y  because [16], by (Select) 
  16] Y >= Y  by (Meta) 

  17] and#(X, or(Y, Z)) >= and#(X, Y)  because and# in Mul, [18] and [19], by (Fun) 
  18] X >= X  by (Meta) 
  19] or(Y, Z) >= Y  because [20], by (Star) 
  20] or*(Y, Z) >= Y  because [21], by (Select) 
  21] Y >= Y  by (Meta) 

  22] and#(X, or(Y, Z)) > and#(X, Z)  because [23], by definition 
  23] and#*(X, or(Y, Z)) >= and#(X, Z)  because and# in Mul, [18] and [24], by (Stat) 
  24] or(Y, Z) > Z  because [25], by definition 
  25] or*(Y, Z) >= Z  because [26], by (Select) 
  26] Z >= Z  by (Meta) 

  27] and#(or(X, Y), Z) >= and#(Z, X)  because and# in Mul, [28] and [31], by (Fun) 
  28] or(X, Y) >= X  because [29], by (Star) 
  29] or*(X, Y) >= X  because [30], by (Select) 
  30] X >= X  by (Meta) 
  31] Z >= Z  by (Meta) 

  32] not(not(X)) >= X  because [33], by (Star) 
  33] not*(not(X)) >= X  because [34], by (Select) 
  34] not(X) >= X  because [35], by (Star) 
  35] not*(X) >= X  because [36], by (Select) 
  36] X >= X  by (Meta) 

  37] not(or(X, Y)) >= and(not(X), not(Y))  because [38], by (Star) 
  38] not*(or(X, Y)) >= and(not(X), not(Y))  because not > and, [39] and [42], by (Copy) 
  39] not*(or(X, Y)) >= not(X)  because not in Mul and [40], by (Stat) 
  40] or(X, Y) > X  because [41], by definition 
  41] or*(X, Y) >= X  because [4], by (Select) 
  42] not*(or(X, Y)) >= not(Y)  because not in Mul and [43], by (Stat) 
  43] or(X, Y) > Y  because [44], by definition 
  44] or*(X, Y) >= Y  because [8], by (Select) 

  45] not(and(X, Y)) >= or(not(X), not(Y))  because [46], by (Star) 
  46] not*(and(X, Y)) >= or(not(X), not(Y))  because not > or, [47] and [50], by (Copy) 
  47] not*(and(X, Y)) >= not(X)  because not in Mul and [48], by (Stat) 
  48] and(X, Y) > X  because [49], by definition 
  49] and*(X, Y) >= X  because [12], by (Select) 
  50] not*(and(X, Y)) >= not(Y)  because not in Mul and [51], by (Stat) 
  51] and(X, Y) > Y  because [52], by definition 
  52] and*(X, Y) >= Y  because [16], by (Select) 

  53] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because [54], by (Star) 
  54] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because and > or, [55] and [58], by (Copy) 
  55] and*(X, or(Y, Z)) >= and(X, Y)  because and in Mul, [18] and [56], by (Stat) 
  56] or(Y, Z) > Y  because [57], by definition 
  57] or*(Y, Z) >= Y  because [21], by (Select) 
  58] and*(X, or(Y, Z)) >= and(X, Z)  because and in Mul, [18] and [24], by (Stat) 

  59] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because [60], by (Star) 
  60] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because and > or, [61] and [64], by (Copy) 
  61] and*(or(X, Y), Z) >= and(Z, X)  because and in Mul, [62] and [31], by (Stat) 
  62] or(X, Y) > X  because [63], by definition 
  63] or*(X, Y) >= X  because [30], by (Select) 
  64] and*(or(X, Y), Z) >= and(Z, Y)  because and in Mul, [65] and [31], by (Stat) 
  65] or(X, Y) > Y  because [66], by definition 
  66] or*(X, Y) >= Y  because [67], by (Select) 
  67] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_2, static, formative) by (P_5, R_2, static, formative), where P_5 consists of:

  not#(or(X, Y)) =#> not#(X)   
  not#(or(X, Y)) =#> not#(Y)   
  not#(and(X, Y)) =#> not#(X)   
  not#(and(X, Y)) =#> not#(Y)   
  and#(X, or(Y, Z)) =#> and#(X, Y)   
  and#(or(X, Y), Z) =#> and#(Z, X)   

Thus, the original system is terminating if (P_5, R_2, static, formative) is finite.

We consider the dependency pair problem (P_5, R_2, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  not#(or(X, Y)) >? not#(X) 
  not#(or(X, Y)) >? not#(Y) 
  not#(and(X, Y)) >? not#(X) 
  not#(and(X, Y)) >? not#(Y) 
  and#(X, or(Y, Z)) >? and#(X, Y) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {and, and#, not, not#, or}, and the following precedence: and# > not > not# > and > or

With these choices, we have:

  1] not#(or(X, Y)) >= not#(X)  because not# in Mul and [2], by (Fun) 
  2] or(X, Y) >= X  because [3], by (Star) 
  3] or*(X, Y) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

  5] not#(or(X, Y)) >= not#(Y)  because not# in Mul and [6], by (Fun) 
  6] or(X, Y) >= Y  because [7], by (Star) 
  7] or*(X, Y) >= Y  because [8], by (Select) 
  8] Y >= Y  by (Meta) 

  9] not#(and(X, Y)) > not#(X)  because [10], by definition 
  10] not#*(and(X, Y)) >= not#(X)  because not# in Mul and [11], by (Stat) 
  11] and(X, Y) > X  because [12], by definition 
  12] and*(X, Y) >= X  because [13], by (Select) 
  13] X >= X  by (Meta) 

  14] not#(and(X, Y)) >= not#(Y)  because not# in Mul and [15], by (Fun) 
  15] and(X, Y) >= Y  because [16], by (Star) 
  16] and*(X, Y) >= Y  because [17], by (Select) 
  17] Y >= Y  by (Meta) 

  18] and#(X, or(Y, Z)) >= and#(X, Y)  because and# in Mul, [19] and [20], by (Fun) 
  19] X >= X  by (Meta) 
  20] or(Y, Z) >= Y  because [21], by (Star) 
  21] or*(Y, Z) >= Y  because [22], by (Select) 
  22] Y >= Y  by (Meta) 

  23] and#(or(X, Y), Z) >= and#(Z, X)  because and# in Mul, [24] and [27], by (Fun) 
  24] or(X, Y) >= X  because [25], by (Star) 
  25] or*(X, Y) >= X  because [26], by (Select) 
  26] X >= X  by (Meta) 
  27] Z >= Z  by (Meta) 

  28] not(not(X)) >= X  because [29], by (Star) 
  29] not*(not(X)) >= X  because [30], by (Select) 
  30] not(X) >= X  because [31], by (Star) 
  31] not*(X) >= X  because [32], by (Select) 
  32] X >= X  by (Meta) 

  33] not(or(X, Y)) >= and(not(X), not(Y))  because [34], by (Star) 
  34] not*(or(X, Y)) >= and(not(X), not(Y))  because not > and, [35] and [38], by (Copy) 
  35] not*(or(X, Y)) >= not(X)  because not in Mul and [36], by (Stat) 
  36] or(X, Y) > X  because [37], by definition 
  37] or*(X, Y) >= X  because [4], by (Select) 
  38] not*(or(X, Y)) >= not(Y)  because not in Mul and [39], by (Stat) 
  39] or(X, Y) > Y  because [40], by definition 
  40] or*(X, Y) >= Y  because [8], by (Select) 

  41] not(and(X, Y)) >= or(not(X), not(Y))  because [42], by (Star) 
  42] not*(and(X, Y)) >= or(not(X), not(Y))  because not > or, [43] and [44], by (Copy) 
  43] not*(and(X, Y)) >= not(X)  because not in Mul and [11], by (Stat) 
  44] not*(and(X, Y)) >= not(Y)  because not in Mul and [45], by (Stat) 
  45] and(X, Y) > Y  because [46], by definition 
  46] and*(X, Y) >= Y  because [17], by (Select) 

  47] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because [48], by (Star) 
  48] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because and > or, [49] and [52], by (Copy) 
  49] and*(X, or(Y, Z)) >= and(X, Y)  because and in Mul, [19] and [50], by (Stat) 
  50] or(Y, Z) > Y  because [51], by definition 
  51] or*(Y, Z) >= Y  because [22], by (Select) 
  52] and*(X, or(Y, Z)) >= and(X, Z)  because and in Mul, [19] and [53], by (Stat) 
  53] or(Y, Z) > Z  because [54], by definition 
  54] or*(Y, Z) >= Z  because [55], by (Select) 
  55] Z >= Z  by (Meta) 

  56] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because [57], by (Star) 
  57] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because and > or, [58] and [61], by (Copy) 
  58] and*(or(X, Y), Z) >= and(Z, X)  because and in Mul, [59] and [27], by (Stat) 
  59] or(X, Y) > X  because [60], by definition 
  60] or*(X, Y) >= X  because [26], by (Select) 
  61] and*(or(X, Y), Z) >= and(Z, Y)  because and in Mul, [62] and [27], by (Stat) 
  62] or(X, Y) > Y  because [63], by definition 
  63] or*(X, Y) >= Y  because [64], by (Select) 
  64] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_2, static, formative) by (P_6, R_2, static, formative), where P_6 consists of:

  not#(or(X, Y)) =#> not#(X)   
  not#(or(X, Y)) =#> not#(Y)   
  not#(and(X, Y)) =#> not#(Y)   
  and#(X, or(Y, Z)) =#> and#(X, Y)   
  and#(or(X, Y), Z) =#> and#(Z, X)   

Thus, the original system is terminating if (P_6, R_2, static, formative) is finite.

We consider the dependency pair problem (P_6, R_2, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  not#(or(X, Y)) >? not#(X) 
  not#(or(X, Y)) >? not#(Y) 
  not#(and(X, Y)) >? not#(Y) 
  and#(X, or(Y, Z)) >? and#(X, Y) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {and, and#, not, not#, or}, and the following precedence: and# > not > and > not# > or

With these choices, we have:

  1] not#(or(X, Y)) >= not#(X)  because not# in Mul and [2], by (Fun) 
  2] or(X, Y) >= X  because [3], by (Star) 
  3] or*(X, Y) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

  5] not#(or(X, Y)) >= not#(Y)  because [6], by (Star) 
  6] not#*(or(X, Y)) >= not#(Y)  because not# in Mul and [7], by (Stat) 
  7] or(X, Y) > Y  because [8], by definition 
  8] or*(X, Y) >= Y  because [9], by (Select) 
  9] Y >= Y  by (Meta) 

  10] not#(and(X, Y)) > not#(Y)  because [11], by definition 
  11] not#*(and(X, Y)) >= not#(Y)  because [12], by (Select) 
  12] and(X, Y) >= not#(Y)  because [13], by (Star) 
  13] and*(X, Y) >= not#(Y)  because and > not# and [14], by (Copy) 
  14] and*(X, Y) >= Y  because [15], by (Select) 
  15] Y >= Y  by (Meta) 

  16] and#(X, or(Y, Z)) >= and#(X, Y)  because and# in Mul, [17] and [18], by (Fun) 
  17] X >= X  by (Meta) 
  18] or(Y, Z) >= Y  because [19], by (Star) 
  19] or*(Y, Z) >= Y  because [20], by (Select) 
  20] Y >= Y  by (Meta) 

  21] and#(or(X, Y), Z) >= and#(Z, X)  because and# in Mul, [22] and [25], by (Fun) 
  22] or(X, Y) >= X  because [23], by (Star) 
  23] or*(X, Y) >= X  because [24], by (Select) 
  24] X >= X  by (Meta) 
  25] Z >= Z  by (Meta) 

  26] not(not(X)) >= X  because [27], by (Star) 
  27] not*(not(X)) >= X  because [28], by (Select) 
  28] not(X) >= X  because [29], by (Star) 
  29] not*(X) >= X  because [30], by (Select) 
  30] X >= X  by (Meta) 

  31] not(or(X, Y)) >= and(not(X), not(Y))  because [32], by (Star) 
  32] not*(or(X, Y)) >= and(not(X), not(Y))  because not > and, [33] and [36], by (Copy) 
  33] not*(or(X, Y)) >= not(X)  because not in Mul and [34], by (Stat) 
  34] or(X, Y) > X  because [35], by definition 
  35] or*(X, Y) >= X  because [4], by (Select) 
  36] not*(or(X, Y)) >= not(Y)  because not in Mul and [7], by (Stat) 

  37] not(and(X, Y)) >= or(not(X), not(Y))  because [38], by (Star) 
  38] not*(and(X, Y)) >= or(not(X), not(Y))  because not > or, [39] and [43], by (Copy) 
  39] not*(and(X, Y)) >= not(X)  because not in Mul and [40], by (Stat) 
  40] and(X, Y) > X  because [41], by definition 
  41] and*(X, Y) >= X  because [42], by (Select) 
  42] X >= X  by (Meta) 
  43] not*(and(X, Y)) >= not(Y)  because not in Mul and [44], by (Stat) 
  44] and(X, Y) > Y  because [14], by definition 

  45] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because [46], by (Star) 
  46] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because and > or, [47] and [50], by (Copy) 
  47] and*(X, or(Y, Z)) >= and(X, Y)  because and in Mul, [17] and [48], by (Stat) 
  48] or(Y, Z) > Y  because [49], by definition 
  49] or*(Y, Z) >= Y  because [20], by (Select) 
  50] and*(X, or(Y, Z)) >= and(X, Z)  because and in Mul, [17] and [51], by (Stat) 
  51] or(Y, Z) > Z  because [52], by definition 
  52] or*(Y, Z) >= Z  because [53], by (Select) 
  53] Z >= Z  by (Meta) 

  54] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because [55], by (Star) 
  55] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because and > or, [56] and [59], by (Copy) 
  56] and*(or(X, Y), Z) >= and(Z, X)  because and in Mul, [57] and [25], by (Stat) 
  57] or(X, Y) > X  because [58], by definition 
  58] or*(X, Y) >= X  because [24], by (Select) 
  59] and*(or(X, Y), Z) >= and(Z, Y)  because and in Mul, [60] and [25], by (Stat) 
  60] or(X, Y) > Y  because [61], by definition 
  61] or*(X, Y) >= Y  because [62], by (Select) 
  62] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_6, R_2, static, formative) by (P_7, R_2, static, formative), where P_7 consists of:

  not#(or(X, Y)) =#> not#(X)   
  not#(or(X, Y)) =#> not#(Y)   
  and#(X, or(Y, Z)) =#> and#(X, Y)   
  and#(or(X, Y), Z) =#> and#(Z, X)   

Thus, the original system is terminating if (P_7, R_2, static, formative) is finite.

We consider the dependency pair problem (P_7, R_2, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  not#(or(X, Y)) >? not#(X) 
  not#(or(X, Y)) >? not#(Y) 
  and#(X, or(Y, Z)) >? and#(X, Y) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {not#} and Mul = {and, and#, not, or}, and the following precedence: and# > not > and > or > not#

With these choices, we have:

  1] not#(or(X, Y)) > not#(X)  because [2], by definition 
  2] not#*(or(X, Y)) >= not#(X)  because [3], by (Select) 
  3] or(X, Y) >= not#(X)  because [4], by (Star) 
  4] or*(X, Y) >= not#(X)  because or > not# and [5], by (Copy) 
  5] or*(X, Y) >= X  because [6], by (Select) 
  6] X >= X  by (Meta) 

  7] not#(or(X, Y)) >= not#(Y)  because [8], by (Fun) 
  8] or(X, Y) >= Y  because [9], by (Star) 
  9] or*(X, Y) >= Y  because [10], by (Select) 
  10] Y >= Y  by (Meta) 

  11] and#(X, or(Y, Z)) >= and#(X, Y)  because and# in Mul, [12] and [13], by (Fun) 
  12] X >= X  by (Meta) 
  13] or(Y, Z) >= Y  because [14], by (Star) 
  14] or*(Y, Z) >= Y  because [15], by (Select) 
  15] Y >= Y  by (Meta) 

  16] and#(or(X, Y), Z) >= and#(Z, X)  because and# in Mul, [17] and [20], by (Fun) 
  17] or(X, Y) >= X  because [18], by (Star) 
  18] or*(X, Y) >= X  because [19], by (Select) 
  19] X >= X  by (Meta) 
  20] Z >= Z  by (Meta) 

  21] not(not(X)) >= X  because [22], by (Star) 
  22] not*(not(X)) >= X  because [23], by (Select) 
  23] not(X) >= X  because [24], by (Star) 
  24] not*(X) >= X  because [25], by (Select) 
  25] X >= X  by (Meta) 

  26] not(or(X, Y)) >= and(not(X), not(Y))  because [27], by (Star) 
  27] not*(or(X, Y)) >= and(not(X), not(Y))  because not > and, [28] and [30], by (Copy) 
  28] not*(or(X, Y)) >= not(X)  because not in Mul and [29], by (Stat) 
  29] or(X, Y) > X  because [5], by definition 
  30] not*(or(X, Y)) >= not(Y)  because not in Mul and [31], by (Stat) 
  31] or(X, Y) > Y  because [32], by definition 
  32] or*(X, Y) >= Y  because [10], by (Select) 

  33] not(and(X, Y)) >= or(not(X), not(Y))  because [34], by (Star) 
  34] not*(and(X, Y)) >= or(not(X), not(Y))  because not > or, [35] and [39], by (Copy) 
  35] not*(and(X, Y)) >= not(X)  because not in Mul and [36], by (Stat) 
  36] and(X, Y) > X  because [37], by definition 
  37] and*(X, Y) >= X  because [38], by (Select) 
  38] X >= X  by (Meta) 
  39] not*(and(X, Y)) >= not(Y)  because not in Mul and [40], by (Stat) 
  40] and(X, Y) > Y  because [41], by definition 
  41] and*(X, Y) >= Y  because [42], by (Select) 
  42] Y >= Y  by (Meta) 

  43] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because [44], by (Star) 
  44] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because and > or, [45] and [48], by (Copy) 
  45] and*(X, or(Y, Z)) >= and(X, Y)  because and in Mul, [12] and [46], by (Stat) 
  46] or(Y, Z) > Y  because [47], by definition 
  47] or*(Y, Z) >= Y  because [15], by (Select) 
  48] and*(X, or(Y, Z)) >= and(X, Z)  because and in Mul, [12] and [49], by (Stat) 
  49] or(Y, Z) > Z  because [50], by definition 
  50] or*(Y, Z) >= Z  because [51], by (Select) 
  51] Z >= Z  by (Meta) 

  52] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because [53], by (Star) 
  53] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because and > or, [54] and [57], by (Copy) 
  54] and*(or(X, Y), Z) >= and(Z, X)  because and in Mul, [55] and [20], by (Stat) 
  55] or(X, Y) > X  because [56], by definition 
  56] or*(X, Y) >= X  because [19], by (Select) 
  57] and*(or(X, Y), Z) >= and(Z, Y)  because and in Mul, [58] and [20], by (Stat) 
  58] or(X, Y) > Y  because [59], by definition 
  59] or*(X, Y) >= Y  because [60], by (Select) 
  60] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_7, R_2, static, formative) by (P_8, R_2, static, formative), where P_8 consists of:

  not#(or(X, Y)) =#> not#(Y)   
  and#(X, or(Y, Z)) =#> and#(X, Y)   
  and#(or(X, Y), Z) =#> and#(Z, X)   

Thus, the original system is terminating if (P_8, R_2, static, formative) is finite.

We consider the dependency pair problem (P_8, R_2, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  not#(or(X, Y)) >? not#(Y) 
  and#(X, or(Y, Z)) >? and#(X, Y) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {and, and#, not, not#, or}, and the following precedence: and# > not > not# > and > or

With these choices, we have:

  1] not#(or(X, Y)) >= not#(Y)  because not# in Mul and [2], by (Fun) 
  2] or(X, Y) >= Y  because [3], by (Star) 
  3] or*(X, Y) >= Y  because [4], by (Select) 
  4] Y >= Y  by (Meta) 

  5] and#(X, or(Y, Z)) > and#(X, Y)  because [6], by definition 
  6] and#*(X, or(Y, Z)) >= and#(X, Y)  because and# in Mul, [7] and [8], by (Stat) 
  7] X >= X  by (Meta) 
  8] or(Y, Z) > Y  because [9], by definition 
  9] or*(Y, Z) >= Y  because [10], by (Select) 
  10] Y >= Y  by (Meta) 

  11] and#(or(X, Y), Z) >= and#(Z, X)  because and# in Mul, [12] and [15], by (Fun) 
  12] or(X, Y) >= X  because [13], by (Star) 
  13] or*(X, Y) >= X  because [14], by (Select) 
  14] X >= X  by (Meta) 
  15] Z >= Z  by (Meta) 

  16] not(not(X)) >= X  because [17], by (Star) 
  17] not*(not(X)) >= X  because [18], by (Select) 
  18] not(X) >= X  because [19], by (Star) 
  19] not*(X) >= X  because [20], by (Select) 
  20] X >= X  by (Meta) 

  21] not(or(X, Y)) >= and(not(X), not(Y))  because [22], by (Star) 
  22] not*(or(X, Y)) >= and(not(X), not(Y))  because not > and, [23] and [27], by (Copy) 
  23] not*(or(X, Y)) >= not(X)  because not in Mul and [24], by (Stat) 
  24] or(X, Y) > X  because [25], by definition 
  25] or*(X, Y) >= X  because [26], by (Select) 
  26] X >= X  by (Meta) 
  27] not*(or(X, Y)) >= not(Y)  because not in Mul and [28], by (Stat) 
  28] or(X, Y) > Y  because [29], by definition 
  29] or*(X, Y) >= Y  because [4], by (Select) 

  30] not(and(X, Y)) >= or(not(X), not(Y))  because [31], by (Star) 
  31] not*(and(X, Y)) >= or(not(X), not(Y))  because not > or, [32] and [36], by (Copy) 
  32] not*(and(X, Y)) >= not(X)  because not in Mul and [33], by (Stat) 
  33] and(X, Y) > X  because [34], by definition 
  34] and*(X, Y) >= X  because [35], by (Select) 
  35] X >= X  by (Meta) 
  36] not*(and(X, Y)) >= not(Y)  because not in Mul and [37], by (Stat) 
  37] and(X, Y) > Y  because [38], by definition 
  38] and*(X, Y) >= Y  because [39], by (Select) 
  39] Y >= Y  by (Meta) 

  40] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because [41], by (Star) 
  41] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because and > or, [42] and [43], by (Copy) 
  42] and*(X, or(Y, Z)) >= and(X, Y)  because and in Mul, [7] and [8], by (Stat) 
  43] and*(X, or(Y, Z)) >= and(X, Z)  because and in Mul, [7] and [44], by (Stat) 
  44] or(Y, Z) > Z  because [45], by definition 
  45] or*(Y, Z) >= Z  because [46], by (Select) 
  46] Z >= Z  by (Meta) 

  47] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because [48], by (Star) 
  48] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because and > or, [49] and [52], by (Copy) 
  49] and*(or(X, Y), Z) >= and(Z, X)  because and in Mul, [50] and [15], by (Stat) 
  50] or(X, Y) > X  because [51], by definition 
  51] or*(X, Y) >= X  because [14], by (Select) 
  52] and*(or(X, Y), Z) >= and(Z, Y)  because and in Mul, [53] and [15], by (Stat) 
  53] or(X, Y) > Y  because [54], by definition 
  54] or*(X, Y) >= Y  because [55], by (Select) 
  55] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_8, R_2, static, formative) by (P_9, R_2, static, formative), where P_9 consists of:

  not#(or(X, Y)) =#> not#(Y)   
  and#(or(X, Y), Z) =#> and#(Z, X)   

Thus, the original system is terminating if (P_9, R_2, static, formative) is finite.

We consider the dependency pair problem (P_9, R_2, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  not#(or(X, Y)) >? not#(Y) 
  and#(or(X, Y), Z) >? and#(Z, X) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[not#(x_1)]] = x_1 

We choose Lex = {} and Mul = {and, and#, not, or}, and the following precedence: and# > not > and > or

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  or(X, Y) > Y 
  and#(or(X, Y), Z) >= and#(Z, X) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) 

With these choices, we have:

  1] or(X, Y) > Y  because [2], by definition 
  2] or*(X, Y) >= Y  because [3], by (Select) 
  3] Y >= Y  by (Meta) 

  4] and#(or(X, Y), Z) >= and#(Z, X)  because and# in Mul, [5] and [8], by (Fun) 
  5] or(X, Y) >= X  because [6], by (Star) 
  6] or*(X, Y) >= X  because [7], by (Select) 
  7] X >= X  by (Meta) 
  8] Z >= Z  by (Meta) 

  9] not(not(X)) >= X  because [10], by (Star) 
  10] not*(not(X)) >= X  because [11], by (Select) 
  11] not(X) >= X  because [12], by (Star) 
  12] not*(X) >= X  because [13], by (Select) 
  13] X >= X  by (Meta) 

  14] not(or(X, Y)) >= and(not(X), not(Y))  because [15], by (Star) 
  15] not*(or(X, Y)) >= and(not(X), not(Y))  because not > and, [16] and [20], by (Copy) 
  16] not*(or(X, Y)) >= not(X)  because not in Mul and [17], by (Stat) 
  17] or(X, Y) > X  because [18], by definition 
  18] or*(X, Y) >= X  because [19], by (Select) 
  19] X >= X  by (Meta) 
  20] not*(or(X, Y)) >= not(Y)  because not in Mul and [21], by (Stat) 
  21] or(X, Y) > Y  because [2], by definition 

  22] not(and(X, Y)) >= or(not(X), not(Y))  because [23], by (Star) 
  23] not*(and(X, Y)) >= or(not(X), not(Y))  because not > or, [24] and [28], by (Copy) 
  24] not*(and(X, Y)) >= not(X)  because not in Mul and [25], by (Stat) 
  25] and(X, Y) > X  because [26], by definition 
  26] and*(X, Y) >= X  because [27], by (Select) 
  27] X >= X  by (Meta) 
  28] not*(and(X, Y)) >= not(Y)  because not in Mul and [29], by (Stat) 
  29] and(X, Y) > Y  because [30], by definition 
  30] and*(X, Y) >= Y  because [31], by (Select) 
  31] Y >= Y  by (Meta) 

  32] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because [33], by (Star) 
  33] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because and > or, [34] and [39], by (Copy) 
  34] and*(X, or(Y, Z)) >= and(X, Y)  because and in Mul, [35] and [36], by (Stat) 
  35] X >= X  by (Meta) 
  36] or(Y, Z) > Y  because [37], by definition 
  37] or*(Y, Z) >= Y  because [38], by (Select) 
  38] Y >= Y  by (Meta) 
  39] and*(X, or(Y, Z)) >= and(X, Z)  because and in Mul, [35] and [40], by (Stat) 
  40] or(Y, Z) > Z  because [41], by definition 
  41] or*(Y, Z) >= Z  because [42], by (Select) 
  42] Z >= Z  by (Meta) 

  43] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because [44], by (Star) 
  44] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because and > or, [45] and [48], by (Copy) 
  45] and*(or(X, Y), Z) >= and(Z, X)  because and in Mul, [46] and [8], by (Stat) 
  46] or(X, Y) > X  because [47], by definition 
  47] or*(X, Y) >= X  because [7], by (Select) 
  48] and*(or(X, Y), Z) >= and(Z, Y)  because and in Mul, [49] and [8], by (Stat) 
  49] or(X, Y) > Y  because [50], by definition 
  50] or*(X, Y) >= Y  because [51], by (Select) 
  51] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_9, R_2, static, formative) by (P_10, R_2, static, formative), where P_10 consists of:

  and#(or(X, Y), Z) =#> and#(Z, X)   

Thus, the original system is terminating if (P_10, R_2, static, formative) is finite.

We consider the dependency pair problem (P_10, R_2, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  and#(or(X, Y), Z) >? and#(Z, X) 
  not(not(X)) >= X 
  not(or(X, Y)) >= and(not(X), not(Y)) 
  not(and(X, Y)) >= or(not(X), not(Y)) 
  and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) 
  and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {and, and#, not, or}, and the following precedence: not > and# > and > or

With these choices, we have:

  1] and#(or(X, Y), Z) > and#(Z, X)  because [2], by definition 
  2] and#*(or(X, Y), Z) >= and#(Z, X)  because and# in Mul, [3] and [6], by (Stat) 
  3] or(X, Y) > X  because [4], by definition 
  4] or*(X, Y) >= X  because [5], by (Select) 
  5] X >= X  by (Meta) 
  6] Z >= Z  by (Meta) 

  7] not(not(X)) >= X  because [8], by (Star) 
  8] not*(not(X)) >= X  because [9], by (Select) 
  9] not(X) >= X  because [10], by (Star) 
  10] not*(X) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 

  12] not(or(X, Y)) >= and(not(X), not(Y))  because [13], by (Star) 
  13] not*(or(X, Y)) >= and(not(X), not(Y))  because not > and, [14] and [18], by (Copy) 
  14] not*(or(X, Y)) >= not(X)  because not in Mul and [15], by (Stat) 
  15] or(X, Y) > X  because [16], by definition 
  16] or*(X, Y) >= X  because [17], by (Select) 
  17] X >= X  by (Meta) 
  18] not*(or(X, Y)) >= not(Y)  because not in Mul and [19], by (Stat) 
  19] or(X, Y) > Y  because [20], by definition 
  20] or*(X, Y) >= Y  because [21], by (Select) 
  21] Y >= Y  by (Meta) 

  22] not(and(X, Y)) >= or(not(X), not(Y))  because [23], by (Star) 
  23] not*(and(X, Y)) >= or(not(X), not(Y))  because not > or, [24] and [28], by (Copy) 
  24] not*(and(X, Y)) >= not(X)  because not in Mul and [25], by (Stat) 
  25] and(X, Y) > X  because [26], by definition 
  26] and*(X, Y) >= X  because [27], by (Select) 
  27] X >= X  by (Meta) 
  28] not*(and(X, Y)) >= not(Y)  because not in Mul and [29], by (Stat) 
  29] and(X, Y) > Y  because [30], by definition 
  30] and*(X, Y) >= Y  because [31], by (Select) 
  31] Y >= Y  by (Meta) 

  32] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because [33], by (Star) 
  33] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z))  because and > or, [34] and [39], by (Copy) 
  34] and*(X, or(Y, Z)) >= and(X, Y)  because and in Mul, [35] and [36], by (Stat) 
  35] X >= X  by (Meta) 
  36] or(Y, Z) > Y  because [37], by definition 
  37] or*(Y, Z) >= Y  because [38], by (Select) 
  38] Y >= Y  by (Meta) 
  39] and*(X, or(Y, Z)) >= and(X, Z)  because and in Mul, [35] and [40], by (Stat) 
  40] or(Y, Z) > Z  because [41], by definition 
  41] or*(Y, Z) >= Z  because [42], by (Select) 
  42] Z >= Z  by (Meta) 

  43] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because [44], by (Star) 
  44] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y))  because and > or, [45] and [46], by (Copy) 
  45] and*(or(X, Y), Z) >= and(Z, X)  because and in Mul, [3] and [6], by (Stat) 
  46] and*(or(X, Y), Z) >= and(Z, Y)  because and in Mul, [47] and [6], by (Stat) 
  47] or(X, Y) > Y  because [48], by definition 
  48] or*(X, Y) >= Y  because [49], by (Select) 
  49] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_10, R_2) by ({}, R_2).  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 +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.