/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

  Alphabet:

    0 : [] --> b 
    cons : [b * c] --> c 
    eq : [b * b] --> a 
    false : [] --> a 
    filter : [b -> a * c] --> c 
    filter2 : [a * b -> a * b * c] --> c 
    if!fac6220min : [a * c] --> b 
    if!fac6220replace : [a * b * b * c] --> c 
    le : [b * b] --> a 
    map : [b -> b * c] --> c 
    min : [c] --> b 
    nil : [] --> c 
    replace : [b * b * c] --> c 
    s : [b] --> b 
    sort : [c] --> c 
    true : [] --> a 

  Rules:

    eq(0, 0) => true 
    eq(0, s(x)) => false 
    eq(s(x), 0) => false 
    eq(s(x), s(y)) => eq(x, y) 
    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    min(cons(0, nil)) => 0 
    min(cons(s(x), nil)) => s(x) 
    min(cons(x, cons(y, z))) => if!fac6220min(le(x, y), cons(x, cons(y, z))) 
    if!fac6220min(true, cons(x, cons(y, z))) => min(cons(x, z)) 
    if!fac6220min(false, cons(x, cons(y, z))) => min(cons(y, z)) 
    replace(x, y, nil) => nil 
    replace(x, y, cons(z, u)) => if!fac6220replace(eq(x, z), x, y, cons(z, u)) 
    if!fac6220replace(true, x, y, cons(z, u)) => cons(y, u) 
    if!fac6220replace(false, x, y, cons(z, u)) => cons(z, replace(x, y, u)) 
    sort(nil) => nil 
    sort(cons(x, y)) => cons(min(cons(x, y)), sort(replace(min(cons(x, y)), x, 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 observe that the rules contain a first-order subset:

  eq(0, 0) => true 
  eq(0, s(X)) => false 
  eq(s(X), 0) => false 
  eq(s(X), s(Y)) => eq(X, Y) 
  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 
  min(cons(0, nil)) => 0 
  min(cons(s(X), nil)) => s(X) 
  min(cons(X, cons(Y, Z))) => if!fac6220min(le(X, Y), cons(X, cons(Y, Z))) 
  if!fac6220min(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) 
  if!fac6220min(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) 
  replace(X, Y, nil) => nil 
  replace(X, Y, cons(Z, U)) => if!fac6220replace(eq(X, Z), X, Y, cons(Z, U)) 
  if!fac6220replace(true, X, Y, cons(Z, U)) => cons(Y, U) 
  if!fac6220replace(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) 
  sort(nil) => nil 
  sort(cons(X, Y)) => cons(min(cons(X, Y)), sort(replace(min(cons(X, Y)), X, Y))) 

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) UsableRulesProof [EQUIVALENT]
 ||         (16) QDP
 ||         (17) QReductionProof [EQUIVALENT]
 ||         (18) QDP
 ||         (19) QDPOrderProof [EQUIVALENT]
 ||         (20) QDP
 ||         (21) DependencyGraphProof [EQUIVALENT]
 ||         (22) TRUE
 ||     (23) QDP
 ||         (24) UsableRulesProof [EQUIVALENT]
 ||         (25) QDP
 ||         (26) QReductionProof [EQUIVALENT]
 ||         (27) QDP
 ||         (28) QDPSizeChangeProof [EQUIVALENT]
 ||         (29) YES
 ||     (30) QDP
 ||         (31) UsableRulesProof [EQUIVALENT]
 ||         (32) QDP
 ||         (33) QReductionProof [EQUIVALENT]
 ||         (34) QDP
 ||         (35) QDPSizeChangeProof [EQUIVALENT]
 ||         (36) YES
 ||     (37) QDP
 ||         (38) UsableRulesProof [EQUIVALENT]
 ||         (39) QDP
 ||         (40) QReductionProof [EQUIVALENT]
 ||         (41) QDP
 ||         (42) QDPOrderProof [EQUIVALENT]
 ||         (43) QDP
 ||         (44) PisEmptyProof [EQUIVALENT]
 ||         (45) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
 || 
 || 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:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (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:
 || 
 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    MIN(cons(%X, cons(%Y, %Z))) -> LE(%X, %Y)
 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> EQ(%X, %Z)
 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
 ||    SORT(cons(%X, %Y)) -> MIN(cons(%X, %Y))
 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
 ||    SORT(cons(%X, %Y)) -> REPLACE(min(cons(%X, %Y)), %X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (5) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 4 less nodes.
 || ----------------------------------------
 || 
 || (6)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (7)
 || 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:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 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:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 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].
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (11)
 || 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.
 || ----------------------------------------
 || 
 || (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:
 || *LE(s(%X), s(%Y)) -> LE(%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:
 || 
 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (15) 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.
 || ----------------------------------------
 || 
 || (16)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
 || 
 || 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)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (17) 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].
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (18)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
 || 
 || 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)
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (19) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
 || 
 || POL( IF!FAC6220MIN_2(x_1, x_2) ) = max{0, 2x_1 + 2x_2 - 2}
 || POL( le_2(x_1, x_2) ) = 2
 || POL( 0 ) = 2
 || POL( true ) = 0
 || POL( s_1(x_1) ) = 2
 || POL( false ) = 0
 || POL( MIN_1(x_1) ) = 2x_1 + 2
 || POL( cons_2(x_1, x_2) ) = x_1 + 2x_2 + 2
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || 
 || ----------------------------------------
 || 
 || (20)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 || 
 || 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)
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (21) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
 || ----------------------------------------
 || 
 || (22)
 || TRUE
 || 
 || ----------------------------------------
 || 
 || (23)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (24) 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.
 || ----------------------------------------
 || 
 || (25)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (26) 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].
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (27)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (28) 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:
 || *EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (29)
 || YES
 || 
 || ----------------------------------------
 || 
 || (30)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (31) 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.
 || ----------------------------------------
 || 
 || (32)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (33) 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].
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (34)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (35) 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:
 || *IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
 || The graph contains the following edges 2 >= 1, 3 >= 2, 4 > 3
 || 
 || 
 || *REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 || The graph contains the following edges 1 >= 2, 2 >= 3, 3 >= 4
 || 
 || 
 || ----------------------------------------
 || 
 || (36)
 || YES
 || 
 || ----------------------------------------
 || 
 || (37)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (38) 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.
 || ----------------------------------------
 || 
 || (39)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (40) 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].
 || 
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (41)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (42) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(SORT(x_1)) = x_1
 ||    POL(cons(x_1, x_2)) = 1 + x_2
 ||    POL(eq(x_1, x_2)) = 0
 ||    POL(false) = 0
 ||    POL(if!fac6220min(x_1, x_2)) = 0
 ||    POL(if!fac6220replace(x_1, x_2, x_3, x_4)) = x_4
 ||    POL(le(x_1, x_2)) = 0
 ||    POL(min(x_1)) = 0
 ||    POL(nil) = 0
 ||    POL(replace(x_1, x_2, x_3)) = x_3
 ||    POL(s(x_1)) = 0
 ||    POL(true) = 0
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 || 
 || 
 || ----------------------------------------
 || 
 || (43)
 || Obligation:
 || Q DP problem:
 || P is empty.
 || The TRS R consists of the following rules:
 || 
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (44) PisEmptyProof (EQUIVALENT)
 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
 || ----------------------------------------
 || 
 || (45)
 || YES
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]).

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

  Dependency Pairs P_0:

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

  Rules R_0:

    eq(0, 0) => true 
    eq(0, s(X)) => false 
    eq(s(X), 0) => false 
    eq(s(X), s(Y)) => eq(X, Y) 
    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    min(cons(0, nil)) => 0 
    min(cons(s(X), nil)) => s(X) 
    min(cons(X, cons(Y, Z))) => if!fac6220min(le(X, Y), cons(X, cons(Y, Z))) 
    if!fac6220min(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) 
    if!fac6220min(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) 
    replace(X, Y, nil) => nil 
    replace(X, Y, cons(Z, U)) => if!fac6220replace(eq(X, Z), X, Y, cons(Z, U)) 
    if!fac6220replace(true, X, Y, cons(Z, U)) => cons(Y, U) 
    if!fac6220replace(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) 
    sort(nil) => nil 
    sort(cons(X, Y)) => cons(min(cons(X, Y)), sort(replace(min(cons(X, Y)), X, Y))) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(F, Y) 

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

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

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 
    * 2 : 1 
    * 3 : 1 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

    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)   

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) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(filter2#) = 4 
  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_3, R_0, minimal, f), where P_3 contains:

  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_1, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, 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 :  
    * 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_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, 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 [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, 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 +++

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