4.15/2.06	YES
4.31/2.09	We consider the system theBenchmark.
4.31/2.09	
4.31/2.09	  Alphabet:
4.31/2.09	
4.31/2.09	    0 : [] --> b 
4.31/2.09	    cons : [b * c] --> c 
4.31/2.09	    eq : [b * b] --> a 
4.31/2.09	    false : [] --> a 
4.31/2.09	    filter : [b -> a * c] --> c 
4.31/2.09	    filter2 : [a * b -> a * b * c] --> c 
4.31/2.09	    if!fac6220min : [a * c] --> b 
4.31/2.09	    if!fac6220replace : [a * b * b * c] --> c 
4.31/2.09	    le : [b * b] --> a 
4.31/2.09	    map : [b -> b * c] --> c 
4.31/2.09	    min : [c] --> b 
4.31/2.09	    nil : [] --> c 
4.31/2.09	    replace : [b * b * c] --> c 
4.31/2.09	    s : [b] --> b 
4.31/2.09	    sort : [c] --> c 
4.31/2.09	    true : [] --> a 
4.31/2.09	
4.31/2.09	  Rules:
4.31/2.09	
4.31/2.09	    eq(0, 0) => true 
4.31/2.09	    eq(0, s(x)) => false 
4.31/2.09	    eq(s(x), 0) => false 
4.31/2.09	    eq(s(x), s(y)) => eq(x, y) 
4.31/2.09	    le(0, x) => true 
4.31/2.09	    le(s(x), 0) => false 
4.31/2.09	    le(s(x), s(y)) => le(x, y) 
4.31/2.09	    min(cons(0, nil)) => 0 
4.31/2.09	    min(cons(s(x), nil)) => s(x) 
4.31/2.09	    min(cons(x, cons(y, z))) => if!fac6220min(le(x, y), cons(x, cons(y, z))) 
4.31/2.09	    if!fac6220min(true, cons(x, cons(y, z))) => min(cons(x, z)) 
4.31/2.09	    if!fac6220min(false, cons(x, cons(y, z))) => min(cons(y, z)) 
4.31/2.09	    replace(x, y, nil) => nil 
4.31/2.09	    replace(x, y, cons(z, u)) => if!fac6220replace(eq(x, z), x, y, cons(z, u)) 
4.31/2.09	    if!fac6220replace(true, x, y, cons(z, u)) => cons(y, u) 
4.31/2.09	    if!fac6220replace(false, x, y, cons(z, u)) => cons(z, replace(x, y, u)) 
4.31/2.09	    sort(nil) => nil 
4.31/2.09	    sort(cons(x, y)) => cons(min(cons(x, y)), sort(replace(min(cons(x, y)), x, y))) 
4.31/2.09	    map(f, nil) => nil 
4.31/2.09	    map(f, cons(x, y)) => cons(f x, map(f, y)) 
4.31/2.09	    filter(f, nil) => nil 
4.31/2.09	    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
4.31/2.09	    filter2(true, f, x, y) => cons(x, filter(f, y)) 
4.31/2.09	    filter2(false, f, x, y) => filter(f, y) 
4.31/2.09	
4.31/2.09	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
4.31/2.09	
4.31/2.09	We observe that the rules contain a first-order subset:
4.31/2.09	
4.31/2.09	  eq(0, 0) => true 
4.31/2.09	  eq(0, s(X)) => false 
4.31/2.09	  eq(s(X), 0) => false 
4.31/2.09	  eq(s(X), s(Y)) => eq(X, Y) 
4.31/2.09	  le(0, X) => true 
4.31/2.09	  le(s(X), 0) => false 
4.31/2.09	  le(s(X), s(Y)) => le(X, Y) 
4.31/2.09	  min(cons(0, nil)) => 0 
4.31/2.09	  min(cons(s(X), nil)) => s(X) 
4.31/2.09	  min(cons(X, cons(Y, Z))) => if!fac6220min(le(X, Y), cons(X, cons(Y, Z))) 
4.31/2.09	  if!fac6220min(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) 
4.31/2.09	  if!fac6220min(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) 
4.31/2.09	  replace(X, Y, nil) => nil 
4.31/2.09	  replace(X, Y, cons(Z, U)) => if!fac6220replace(eq(X, Z), X, Y, cons(Z, U)) 
4.31/2.09	  if!fac6220replace(true, X, Y, cons(Z, U)) => cons(Y, U) 
4.31/2.09	  if!fac6220replace(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) 
4.31/2.09	  sort(nil) => nil 
4.31/2.09	  sort(cons(X, Y)) => cons(min(cons(X, Y)), sort(replace(min(cons(X, Y)), X, Y))) 
4.31/2.09	
4.31/2.09	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.
4.31/2.09	
4.31/2.09	According to the external first-order termination prover, this system is indeed terminating:
4.31/2.09	
4.31/2.09	 || proof of resources/system.trs
4.31/2.09	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || Termination w.r.t. Q of the given QTRS could be proven:
4.31/2.09	 || 
4.31/2.09	 || (0) QTRS
4.31/2.09	 || (1) Overlay + Local Confluence [EQUIVALENT]
4.31/2.09	 || (2) QTRS
4.31/2.09	 || (3) DependencyPairsProof [EQUIVALENT]
4.31/2.09	 || (4) QDP
4.31/2.09	 || (5) DependencyGraphProof [EQUIVALENT]
4.31/2.09	 || (6) AND
4.31/2.09	 ||     (7) QDP
4.31/2.09	 ||         (8) UsableRulesProof [EQUIVALENT]
4.31/2.09	 ||         (9) QDP
4.31/2.09	 ||         (10) QReductionProof [EQUIVALENT]
4.31/2.09	 ||         (11) QDP
4.31/2.09	 ||         (12) QDPSizeChangeProof [EQUIVALENT]
4.31/2.09	 ||         (13) YES
4.31/2.09	 ||     (14) QDP
4.31/2.09	 ||         (15) UsableRulesProof [EQUIVALENT]
4.31/2.09	 ||         (16) QDP
4.31/2.09	 ||         (17) QReductionProof [EQUIVALENT]
4.31/2.09	 ||         (18) QDP
4.31/2.09	 ||         (19) QDPOrderProof [EQUIVALENT]
4.31/2.09	 ||         (20) QDP
4.31/2.09	 ||         (21) PisEmptyProof [EQUIVALENT]
4.31/2.09	 ||         (22) YES
4.31/2.09	 ||     (23) QDP
4.31/2.09	 ||         (24) UsableRulesProof [EQUIVALENT]
4.31/2.09	 ||         (25) QDP
4.31/2.09	 ||         (26) QReductionProof [EQUIVALENT]
4.31/2.09	 ||         (27) QDP
4.31/2.09	 ||         (28) QDPSizeChangeProof [EQUIVALENT]
4.31/2.09	 ||         (29) YES
4.31/2.09	 ||     (30) QDP
4.31/2.09	 ||         (31) UsableRulesProof [EQUIVALENT]
4.31/2.09	 ||         (32) QDP
4.31/2.09	 ||         (33) QReductionProof [EQUIVALENT]
4.31/2.09	 ||         (34) QDP
4.31/2.09	 ||         (35) QDPSizeChangeProof [EQUIVALENT]
4.31/2.09	 ||         (36) YES
4.31/2.09	 ||     (37) QDP
4.31/2.09	 ||         (38) UsableRulesProof [EQUIVALENT]
4.31/2.09	 ||         (39) QDP
4.31/2.09	 ||         (40) QReductionProof [EQUIVALENT]
4.31/2.09	 ||         (41) QDP
4.31/2.09	 ||         (42) QDPOrderProof [EQUIVALENT]
4.31/2.09	 ||         (43) QDP
4.31/2.09	 ||         (44) PisEmptyProof [EQUIVALENT]
4.31/2.09	 ||         (45) YES
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (0)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q restricted rewrite system:
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    sort(nil) -> nil
4.31/2.09	 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
4.31/2.09	 || 
4.31/2.09	 || Q is empty.
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (1) Overlay + Local Confluence (EQUIVALENT)
4.31/2.09	 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (2)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q restricted rewrite system:
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    sort(nil) -> nil
4.31/2.09	 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (3) DependencyPairsProof (EQUIVALENT)
4.31/2.09	 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (4)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
4.31/2.09	 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
4.31/2.09	 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    MIN(cons(%X, cons(%Y, %Z))) -> LE(%X, %Y)
4.31/2.09	 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
4.31/2.09	 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
4.31/2.09	 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> EQ(%X, %Z)
4.31/2.09	 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
4.31/2.09	 ||    SORT(cons(%X, %Y)) -> MIN(cons(%X, %Y))
4.31/2.09	 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
4.31/2.09	 ||    SORT(cons(%X, %Y)) -> REPLACE(min(cons(%X, %Y)), %X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    sort(nil) -> nil
4.31/2.09	 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (5) DependencyGraphProof (EQUIVALENT)
4.31/2.09	 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 4 less nodes.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (6)
4.31/2.09	 || Complex Obligation (AND)
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (7)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    sort(nil) -> nil
4.31/2.09	 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (8) UsableRulesProof (EQUIVALENT)
4.31/2.09	 || 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.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (9)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || R is empty.
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (10) QReductionProof (EQUIVALENT)
4.31/2.09	 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (11)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || R is empty.
4.31/2.09	 || Q is empty.
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (12) QDPSizeChangeProof (EQUIVALENT)
4.31/2.09	 || 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. 
4.31/2.09	 || 
4.31/2.09	 || From the DPs we obtained the following set of size-change graphs:
4.31/2.09	 || *LE(s(%X), s(%Y)) -> LE(%X, %Y)
4.31/2.09	 || The graph contains the following edges 1 > 1, 2 > 2
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (13)
4.31/2.09	 || YES
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (14)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
4.31/2.09	 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    sort(nil) -> nil
4.31/2.09	 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (15) UsableRulesProof (EQUIVALENT)
4.31/2.09	 || 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.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (16)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
4.31/2.09	 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (17) QReductionProof (EQUIVALENT)
4.31/2.09	 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (18)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
4.31/2.09	 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (19) QDPOrderProof (EQUIVALENT)
4.31/2.09	 || We use the reduction pair processor [LPAR04,JAR06].
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || The following pairs can be oriented strictly and are deleted.
4.31/2.09	 || 
4.31/2.09	 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
4.31/2.09	 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
4.31/2.09	 || The remaining pairs can at least be oriented weakly.
4.31/2.09	 || Used ordering:  Combined order from the following AFS and order.
4.31/2.09	 || MIN(x1)  =  MIN(x1)
4.31/2.09	 || 
4.31/2.09	 || cons(x1, x2)  =  cons(x2)
4.31/2.09	 || 
4.31/2.09	 || IF!FAC6220MIN(x1, x2)  =  IF!FAC6220MIN(x1, x2)
4.31/2.09	 || 
4.31/2.09	 || le(x1, x2)  =  le
4.31/2.09	 || 
4.31/2.09	 || true  =  true
4.31/2.09	 || 
4.31/2.09	 || false  =  false
4.31/2.09	 || 
4.31/2.09	 || 0  =  0
4.31/2.09	 || 
4.31/2.09	 || s(x1)  =  s
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || Recursive path order with status [RPO].
4.31/2.09	 || Quasi-Precedence: [cons_1, le, true] > [MIN_1, false] > IF!FAC6220MIN_2
4.31/2.09	 || 0 > IF!FAC6220MIN_2
4.31/2.09	 || s > IF!FAC6220MIN_2
4.31/2.09	 || 
4.31/2.09	 || Status: MIN_1: [1]
4.31/2.09	 || cons_1: multiset status
4.31/2.09	 || IF!FAC6220MIN_2: multiset status
4.31/2.09	 || le: multiset status
4.31/2.09	 || true: multiset status
4.31/2.09	 || false: multiset status
4.31/2.09	 || 0: multiset status
4.31/2.09	 || s: []
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
4.31/2.09	 || 
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (20)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || P is empty.
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (21) PisEmptyProof (EQUIVALENT)
4.31/2.09	 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (22)
4.31/2.09	 || YES
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (23)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    sort(nil) -> nil
4.31/2.09	 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (24) UsableRulesProof (EQUIVALENT)
4.31/2.09	 || 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.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (25)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || R is empty.
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (26) QReductionProof (EQUIVALENT)
4.31/2.09	 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (27)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || R is empty.
4.31/2.09	 || Q is empty.
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (28) QDPSizeChangeProof (EQUIVALENT)
4.31/2.09	 || 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. 
4.31/2.09	 || 
4.31/2.09	 || From the DPs we obtained the following set of size-change graphs:
4.31/2.09	 || *EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
4.31/2.09	 || The graph contains the following edges 1 > 1, 2 > 2
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (29)
4.31/2.09	 || YES
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (30)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    sort(nil) -> nil
4.31/2.09	 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (31) UsableRulesProof (EQUIVALENT)
4.31/2.09	 || 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.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (32)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (33) QReductionProof (EQUIVALENT)
4.31/2.09	 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
4.31/2.09	 || 
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (34)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (35) QDPSizeChangeProof (EQUIVALENT)
4.31/2.09	 || 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. 
4.31/2.09	 || 
4.31/2.09	 || From the DPs we obtained the following set of size-change graphs:
4.31/2.09	 || *IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
4.31/2.09	 || The graph contains the following edges 2 >= 1, 3 >= 2, 4 > 3
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || *REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 || The graph contains the following edges 1 >= 2, 2 >= 3, 3 >= 4
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (36)
4.31/2.09	 || YES
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (37)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    sort(nil) -> nil
4.31/2.09	 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (38) UsableRulesProof (EQUIVALENT)
4.31/2.09	 || 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.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (39)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (40) QReductionProof (EQUIVALENT)
4.31/2.09	 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
4.31/2.09	 || 
4.31/2.09	 ||    sort(nil)
4.31/2.09	 ||    sort(cons(x0, x1))
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (41)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || The TRS P consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
4.31/2.09	 || 
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (42) QDPOrderProof (EQUIVALENT)
4.31/2.09	 || We use the reduction pair processor [LPAR04,JAR06].
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || The following pairs can be oriented strictly and are deleted.
4.31/2.09	 || 
4.31/2.09	 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
4.31/2.09	 || The remaining pairs can at least be oriented weakly.
4.31/2.09	 || Used ordering:  Combined order from the following AFS and order.
4.31/2.09	 || SORT(x1)  =  x1
4.31/2.09	 || 
4.31/2.09	 || cons(x1, x2)  =  cons(x2)
4.31/2.09	 || 
4.31/2.09	 || replace(x1, x2, x3)  =  x3
4.31/2.09	 || 
4.31/2.09	 || min(x1)  =  x1
4.31/2.09	 || 
4.31/2.09	 || 0  =  0
4.31/2.09	 || 
4.31/2.09	 || nil  =  nil
4.31/2.09	 || 
4.31/2.09	 || s(x1)  =  s
4.31/2.09	 || 
4.31/2.09	 || if!fac6220min(x1, x2)  =  if!fac6220min(x1, x2)
4.31/2.09	 || 
4.31/2.09	 || le(x1, x2)  =  x2
4.31/2.09	 || 
4.31/2.09	 || true  =  true
4.31/2.09	 || 
4.31/2.09	 || false  =  false
4.31/2.09	 || 
4.31/2.09	 || if!fac6220replace(x1, x2, x3, x4)  =  x4
4.31/2.09	 || 
4.31/2.09	 || eq(x1, x2)  =  eq(x1, x2)
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || Recursive path order with status [RPO].
4.31/2.09	 || Quasi-Precedence: 0 > false
4.31/2.09	 || [s, eq_2] > false
4.31/2.09	 || if!fac6220min_2 > cons_1
4.31/2.09	 || true > cons_1
4.31/2.09	 || 
4.31/2.09	 || Status: cons_1: [1]
4.31/2.09	 || 0: multiset status
4.31/2.09	 || nil: multiset status
4.31/2.09	 || s: multiset status
4.31/2.09	 || if!fac6220min_2: multiset status
4.31/2.09	 || true: multiset status
4.31/2.09	 || false: multiset status
4.31/2.09	 || eq_2: multiset status
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
4.31/2.09	 || 
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 || 
4.31/2.09	 || 
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (43)
4.31/2.09	 || Obligation:
4.31/2.09	 || Q DP problem:
4.31/2.09	 || P is empty.
4.31/2.09	 || The TRS R consists of the following rules:
4.31/2.09	 || 
4.31/2.09	 ||    min(cons(0, nil)) -> 0
4.31/2.09	 ||    min(cons(s(%X), nil)) -> s(%X)
4.31/2.09	 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
4.31/2.09	 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
4.31/2.09	 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
4.31/2.09	 ||    replace(%X, %Y, nil) -> nil
4.31/2.09	 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
4.31/2.09	 ||    eq(0, 0) -> true
4.31/2.09	 ||    eq(0, s(%X)) -> false
4.31/2.09	 ||    eq(s(%X), 0) -> false
4.31/2.09	 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
4.31/2.09	 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
4.31/2.09	 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
4.31/2.09	 ||    le(0, %X) -> true
4.31/2.09	 ||    le(s(%X), 0) -> false
4.31/2.09	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
4.31/2.09	 || 
4.31/2.09	 || The set Q consists of the following terms:
4.31/2.09	 || 
4.31/2.09	 ||    eq(0, 0)
4.31/2.09	 ||    eq(0, s(x0))
4.31/2.09	 ||    eq(s(x0), 0)
4.31/2.09	 ||    eq(s(x0), s(x1))
4.31/2.09	 ||    le(0, x0)
4.31/2.09	 ||    le(s(x0), 0)
4.31/2.09	 ||    le(s(x0), s(x1))
4.31/2.09	 ||    min(cons(0, nil))
4.31/2.09	 ||    min(cons(s(x0), nil))
4.31/2.09	 ||    min(cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
4.31/2.09	 ||    replace(x0, x1, nil)
4.31/2.09	 ||    replace(x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
4.31/2.09	 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
4.31/2.09	 || 
4.31/2.09	 || We have to consider all minimal (P,Q,R)-chains.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (44) PisEmptyProof (EQUIVALENT)
4.31/2.09	 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
4.31/2.09	 || ----------------------------------------
4.31/2.09	 || 
4.31/2.09	 || (45)
4.31/2.09	 || YES
4.31/2.09	 || 
4.31/2.09	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]).
4.31/2.09	
4.31/2.09	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
4.31/2.09	
4.31/2.09	  Dependency Pairs P_0:
4.31/2.09	
4.31/2.09	    0] map#(F, cons(X, Y)) =#> map#(F, Y)   
4.31/2.09	    1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
4.31/2.09	    2] filter2#(true, F, X, Y) =#> filter#(F, Y)   
4.31/2.09	    3] filter2#(false, F, X, Y) =#> filter#(F, Y)   
4.31/2.09	
4.31/2.09	  Rules R_0:
4.31/2.09	
4.31/2.09	    eq(0, 0) => true 
4.31/2.09	    eq(0, s(X)) => false 
4.31/2.09	    eq(s(X), 0) => false 
4.31/2.09	    eq(s(X), s(Y)) => eq(X, Y) 
4.31/2.09	    le(0, X) => true 
4.31/2.09	    le(s(X), 0) => false 
4.31/2.09	    le(s(X), s(Y)) => le(X, Y) 
4.31/2.09	    min(cons(0, nil)) => 0 
4.31/2.09	    min(cons(s(X), nil)) => s(X) 
4.31/2.09	    min(cons(X, cons(Y, Z))) => if!fac6220min(le(X, Y), cons(X, cons(Y, Z))) 
4.31/2.09	    if!fac6220min(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) 
4.31/2.09	    if!fac6220min(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) 
4.31/2.09	    replace(X, Y, nil) => nil 
4.31/2.09	    replace(X, Y, cons(Z, U)) => if!fac6220replace(eq(X, Z), X, Y, cons(Z, U)) 
4.31/2.09	    if!fac6220replace(true, X, Y, cons(Z, U)) => cons(Y, U) 
4.31/2.09	    if!fac6220replace(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) 
4.31/2.09	    sort(nil) => nil 
4.31/2.09	    sort(cons(X, Y)) => cons(min(cons(X, Y)), sort(replace(min(cons(X, Y)), X, Y))) 
4.31/2.09	    map(F, nil) => nil 
4.31/2.09	    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
4.31/2.09	    filter(F, nil) => nil 
4.31/2.09	    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
4.31/2.09	    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
4.31/2.09	    filter2(false, F, X, Y) => filter(F, Y) 
4.31/2.09	
4.31/2.09	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
4.31/2.09	
4.31/2.09	We consider the dependency pair problem (P_0, R_0, static, formative).
4.31/2.09	
4.31/2.09	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
4.31/2.09	
4.31/2.09	    * 0 : 0 
4.31/2.09	    * 1 : 2, 3 
4.31/2.09	    * 2 : 1 
4.31/2.09	    * 3 : 1 
4.31/2.09	
4.31/2.09	This graph has the following strongly connected components:
4.31/2.09	
4.31/2.09	  P_1:
4.31/2.09	
4.31/2.09	    map#(F, cons(X, Y)) =#> map#(F, Y)   
4.31/2.09	
4.31/2.09	  P_2:
4.31/2.09	
4.31/2.09	    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
4.31/2.09	    filter2#(true, F, X, Y) =#> filter#(F, Y)   
4.31/2.09	    filter2#(false, F, X, Y) =#> filter#(F, Y)   
4.31/2.09	
4.31/2.09	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).
4.31/2.09	
4.31/2.09	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.
4.31/2.09	
4.31/2.09	We consider the dependency pair problem (P_2, R_0, static, formative).
4.31/2.09	
4.31/2.09	We apply the subterm criterion with the following projection function:
4.31/2.09	
4.31/2.09	  nu(filter2#) = 4 
4.31/2.09	  nu(filter#) = 2 
4.31/2.09	
4.31/2.09	Thus, we can orient the dependency pairs as follows:
4.31/2.09	
4.31/2.09	  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
4.31/2.09	  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
4.31/2.09	  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
4.31/2.09	
4.31/2.09	By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_2, R_0, static, f) by (P_3, R_0, static, f), where P_3 contains:
4.31/2.09	
4.31/2.09	  filter2#(true, F, X, Y) =#> filter#(F, Y)   
4.31/2.09	  filter2#(false, F, X, Y) =#> filter#(F, Y)   
4.31/2.09	
4.31/2.09	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite.
4.31/2.09	
4.31/2.09	We consider the dependency pair problem (P_3, R_0, static, formative).
4.31/2.09	
4.31/2.09	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
4.31/2.09	
4.31/2.09	    * 0 :  
4.31/2.09	    * 1 :  
4.31/2.09	
4.31/2.09	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
4.31/2.09	
4.31/2.09	Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.
4.31/2.09	
4.31/2.09	We consider the dependency pair problem (P_1, R_0, static, formative).
4.31/2.09	
4.31/2.09	We apply the subterm criterion with the following projection function:
4.31/2.09	
4.31/2.09	  nu(map#) = 2 
4.31/2.09	
4.31/2.09	Thus, we can orient the dependency pairs as follows:
4.31/2.09	
4.31/2.09	  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 
4.31/2.09	
4.31/2.09	By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_1, R_0, static, f) by ({}, R_0, static, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.
4.31/2.09	
4.31/2.09	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
4.31/2.09	
4.31/2.09	
4.31/2.09	+++ Citations +++
4.31/2.09	
4.31/2.09	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
4.31/2.09	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
4.31/2.09	[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.
4.31/2.09	EOF