2.57/1.30	YES
2.57/1.31	We consider the system theBenchmark.
2.57/1.31	
2.57/1.31	  Alphabet:
2.57/1.31	
2.57/1.31	    0 : [] --> c 
2.57/1.31	    cons : [c * d] --> d 
2.57/1.31	    false : [] --> b 
2.57/1.31	    filter : [c -> b * d] --> d 
2.57/1.31	    filter2 : [b * c -> b * c * d] --> d 
2.57/1.31	    map : [c -> c * d] --> d 
2.57/1.31	    nil : [] --> d 
2.57/1.31	    rev : [d] --> d 
2.57/1.31	    rev1 : [c * d] --> c 
2.57/1.31	    rev2 : [c * d] --> d 
2.57/1.31	    s : [a] --> c 
2.57/1.31	    true : [] --> b 
2.57/1.31	
2.57/1.31	  Rules:
2.57/1.31	
2.57/1.31	    rev(nil) => nil 
2.57/1.31	    rev(cons(x, y)) => cons(rev1(x, y), rev2(x, y)) 
2.57/1.31	    rev1(0, nil) => 0 
2.57/1.31	    rev1(s(x), nil) => s(x) 
2.57/1.31	    rev1(x, cons(y, z)) => rev1(y, z) 
2.57/1.31	    rev2(x, nil) => nil 
2.57/1.31	    rev2(x, cons(y, z)) => rev(cons(x, rev2(y, z))) 
2.57/1.31	    map(f, nil) => nil 
2.57/1.31	    map(f, cons(x, y)) => cons(f x, map(f, y)) 
2.57/1.31	    filter(f, nil) => nil 
2.57/1.31	    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
2.57/1.31	    filter2(true, f, x, y) => cons(x, filter(f, y)) 
2.57/1.31	    filter2(false, f, x, y) => filter(f, y) 
2.57/1.31	
2.57/1.31	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
2.57/1.31	
2.57/1.31	We observe that the rules contain a first-order subset:
2.57/1.31	
2.57/1.31	  rev(nil) => nil 
2.57/1.31	  rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) 
2.57/1.31	  rev1(0, nil) => 0 
2.57/1.31	  rev1(s(X), nil) => s(X) 
2.57/1.31	  rev1(X, cons(Y, Z)) => rev1(Y, Z) 
2.57/1.31	  rev2(X, nil) => nil 
2.57/1.31	  rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) 
2.57/1.31	
2.57/1.31	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.
2.57/1.31	
2.57/1.31	According to the external first-order termination prover, this system is indeed terminating:
2.57/1.31	
2.57/1.31	 || proof of resources/system.trs
2.57/1.31	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
2.57/1.31	 || 
2.57/1.31	 || 
2.57/1.31	 || Termination w.r.t. Q of the given QTRS could be proven:
2.57/1.31	 || 
2.57/1.31	 || (0) QTRS
2.57/1.31	 || (1) Overlay + Local Confluence [EQUIVALENT]
2.57/1.31	 || (2) QTRS
2.57/1.31	 || (3) DependencyPairsProof [EQUIVALENT]
2.57/1.31	 || (4) QDP
2.57/1.31	 || (5) DependencyGraphProof [EQUIVALENT]
2.57/1.31	 || (6) AND
2.57/1.31	 ||     (7) QDP
2.57/1.31	 ||         (8) UsableRulesProof [EQUIVALENT]
2.57/1.31	 ||         (9) QDP
2.57/1.31	 ||         (10) QReductionProof [EQUIVALENT]
2.57/1.31	 ||         (11) QDP
2.57/1.31	 ||         (12) QDPSizeChangeProof [EQUIVALENT]
2.57/1.31	 ||         (13) YES
2.57/1.31	 ||     (14) QDP
2.57/1.31	 ||         (15) UsableRulesProof [EQUIVALENT]
2.57/1.31	 ||         (16) QDP
2.57/1.31	 ||         (17) QDPOrderProof [EQUIVALENT]
2.57/1.31	 ||         (18) QDP
2.57/1.31	 ||         (19) DependencyGraphProof [EQUIVALENT]
2.57/1.31	 ||         (20) TRUE
2.57/1.31	 || 
2.57/1.31	 || 
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (0)
2.57/1.31	 || Obligation:
2.57/1.31	 || Q restricted rewrite system:
2.57/1.31	 || The TRS R consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil) -> nil
2.57/1.31	 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
2.57/1.31	 ||    rev1(0, nil) -> 0
2.57/1.31	 ||    rev1(s(%X), nil) -> s(%X)
2.57/1.31	 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
2.57/1.31	 ||    rev2(%X, nil) -> nil
2.57/1.31	 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 || 
2.57/1.31	 || Q is empty.
2.57/1.31	 || 
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (1) Overlay + Local Confluence (EQUIVALENT)
2.57/1.31	 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (2)
2.57/1.31	 || Obligation:
2.57/1.31	 || Q restricted rewrite system:
2.57/1.31	 || The TRS R consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil) -> nil
2.57/1.31	 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
2.57/1.31	 ||    rev1(0, nil) -> 0
2.57/1.31	 ||    rev1(s(%X), nil) -> s(%X)
2.57/1.31	 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
2.57/1.31	 ||    rev2(%X, nil) -> nil
2.57/1.31	 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 || 
2.57/1.31	 || The set Q consists of the following terms:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil)
2.57/1.31	 ||    rev(cons(x0, x1))
2.57/1.31	 ||    rev1(0, nil)
2.57/1.31	 ||    rev1(s(x0), nil)
2.57/1.31	 ||    rev1(x0, cons(x1, x2))
2.57/1.31	 ||    rev2(x0, nil)
2.57/1.31	 ||    rev2(x0, cons(x1, x2))
2.57/1.31	 || 
2.57/1.31	 || 
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (3) DependencyPairsProof (EQUIVALENT)
2.57/1.31	 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (4)
2.57/1.31	 || Obligation:
2.57/1.31	 || Q DP problem:
2.57/1.31	 || The TRS P consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    REV(cons(%X, %Y)) -> REV1(%X, %Y)
2.57/1.31	 ||    REV(cons(%X, %Y)) -> REV2(%X, %Y)
2.57/1.31	 ||    REV1(%X, cons(%Y, %Z)) -> REV1(%Y, %Z)
2.57/1.31	 ||    REV2(%X, cons(%Y, %Z)) -> REV(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 ||    REV2(%X, cons(%Y, %Z)) -> REV2(%Y, %Z)
2.57/1.31	 || 
2.57/1.31	 || The TRS R consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil) -> nil
2.57/1.31	 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
2.57/1.31	 ||    rev1(0, nil) -> 0
2.57/1.31	 ||    rev1(s(%X), nil) -> s(%X)
2.57/1.31	 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
2.57/1.31	 ||    rev2(%X, nil) -> nil
2.57/1.31	 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 || 
2.57/1.31	 || The set Q consists of the following terms:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil)
2.57/1.31	 ||    rev(cons(x0, x1))
2.57/1.31	 ||    rev1(0, nil)
2.57/1.31	 ||    rev1(s(x0), nil)
2.57/1.31	 ||    rev1(x0, cons(x1, x2))
2.57/1.31	 ||    rev2(x0, nil)
2.57/1.31	 ||    rev2(x0, cons(x1, x2))
2.57/1.31	 || 
2.57/1.31	 || We have to consider all minimal (P,Q,R)-chains.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (5) DependencyGraphProof (EQUIVALENT)
2.57/1.31	 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (6)
2.57/1.31	 || Complex Obligation (AND)
2.57/1.31	 || 
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (7)
2.57/1.31	 || Obligation:
2.57/1.31	 || Q DP problem:
2.57/1.31	 || The TRS P consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    REV1(%X, cons(%Y, %Z)) -> REV1(%Y, %Z)
2.57/1.31	 || 
2.57/1.31	 || The TRS R consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil) -> nil
2.57/1.31	 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
2.57/1.31	 ||    rev1(0, nil) -> 0
2.57/1.31	 ||    rev1(s(%X), nil) -> s(%X)
2.57/1.31	 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
2.57/1.31	 ||    rev2(%X, nil) -> nil
2.57/1.31	 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 || 
2.57/1.31	 || The set Q consists of the following terms:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil)
2.57/1.31	 ||    rev(cons(x0, x1))
2.57/1.31	 ||    rev1(0, nil)
2.57/1.31	 ||    rev1(s(x0), nil)
2.57/1.31	 ||    rev1(x0, cons(x1, x2))
2.57/1.31	 ||    rev2(x0, nil)
2.57/1.31	 ||    rev2(x0, cons(x1, x2))
2.57/1.31	 || 
2.57/1.31	 || We have to consider all minimal (P,Q,R)-chains.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (8) UsableRulesProof (EQUIVALENT)
2.57/1.31	 || 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.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (9)
2.57/1.31	 || Obligation:
2.57/1.31	 || Q DP problem:
2.57/1.31	 || The TRS P consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    REV1(%X, cons(%Y, %Z)) -> REV1(%Y, %Z)
2.57/1.31	 || 
2.57/1.31	 || R is empty.
2.57/1.31	 || The set Q consists of the following terms:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil)
2.57/1.31	 ||    rev(cons(x0, x1))
2.57/1.31	 ||    rev1(0, nil)
2.57/1.31	 ||    rev1(s(x0), nil)
2.57/1.31	 ||    rev1(x0, cons(x1, x2))
2.57/1.31	 ||    rev2(x0, nil)
2.57/1.31	 ||    rev2(x0, cons(x1, x2))
2.57/1.31	 || 
2.57/1.31	 || We have to consider all minimal (P,Q,R)-chains.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (10) QReductionProof (EQUIVALENT)
2.57/1.31	 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil)
2.57/1.31	 ||    rev(cons(x0, x1))
2.57/1.31	 ||    rev1(0, nil)
2.57/1.31	 ||    rev1(s(x0), nil)
2.57/1.31	 ||    rev1(x0, cons(x1, x2))
2.57/1.31	 ||    rev2(x0, nil)
2.57/1.31	 ||    rev2(x0, cons(x1, x2))
2.57/1.31	 || 
2.57/1.31	 || 
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (11)
2.57/1.31	 || Obligation:
2.57/1.31	 || Q DP problem:
2.57/1.31	 || The TRS P consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    REV1(%X, cons(%Y, %Z)) -> REV1(%Y, %Z)
2.57/1.31	 || 
2.57/1.31	 || R is empty.
2.57/1.31	 || Q is empty.
2.57/1.31	 || We have to consider all minimal (P,Q,R)-chains.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (12) QDPSizeChangeProof (EQUIVALENT)
2.57/1.31	 || 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. 
2.57/1.31	 || 
2.57/1.31	 || From the DPs we obtained the following set of size-change graphs:
2.57/1.31	 || *REV1(%X, cons(%Y, %Z)) -> REV1(%Y, %Z)
2.57/1.31	 || The graph contains the following edges 2 > 1, 2 > 2
2.57/1.31	 || 
2.57/1.31	 || 
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (13)
2.57/1.31	 || YES
2.57/1.31	 || 
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (14)
2.57/1.31	 || Obligation:
2.57/1.31	 || Q DP problem:
2.57/1.31	 || The TRS P consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    REV(cons(%X, %Y)) -> REV2(%X, %Y)
2.57/1.31	 ||    REV2(%X, cons(%Y, %Z)) -> REV(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 ||    REV2(%X, cons(%Y, %Z)) -> REV2(%Y, %Z)
2.57/1.31	 || 
2.57/1.31	 || The TRS R consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil) -> nil
2.57/1.31	 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
2.57/1.31	 ||    rev1(0, nil) -> 0
2.57/1.31	 ||    rev1(s(%X), nil) -> s(%X)
2.57/1.31	 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
2.57/1.31	 ||    rev2(%X, nil) -> nil
2.57/1.31	 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 || 
2.57/1.31	 || The set Q consists of the following terms:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil)
2.57/1.31	 ||    rev(cons(x0, x1))
2.57/1.31	 ||    rev1(0, nil)
2.57/1.31	 ||    rev1(s(x0), nil)
2.57/1.31	 ||    rev1(x0, cons(x1, x2))
2.57/1.31	 ||    rev2(x0, nil)
2.57/1.31	 ||    rev2(x0, cons(x1, x2))
2.57/1.31	 || 
2.57/1.31	 || We have to consider all minimal (P,Q,R)-chains.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (15) UsableRulesProof (EQUIVALENT)
2.57/1.31	 || 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.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (16)
2.57/1.31	 || Obligation:
2.57/1.31	 || Q DP problem:
2.57/1.31	 || The TRS P consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    REV(cons(%X, %Y)) -> REV2(%X, %Y)
2.57/1.31	 ||    REV2(%X, cons(%Y, %Z)) -> REV(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 ||    REV2(%X, cons(%Y, %Z)) -> REV2(%Y, %Z)
2.57/1.31	 || 
2.57/1.31	 || The TRS R consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    rev2(%X, nil) -> nil
2.57/1.31	 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
2.57/1.31	 ||    rev1(0, nil) -> 0
2.57/1.31	 ||    rev1(s(%X), nil) -> s(%X)
2.57/1.31	 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
2.57/1.31	 || 
2.57/1.31	 || The set Q consists of the following terms:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil)
2.57/1.31	 ||    rev(cons(x0, x1))
2.57/1.31	 ||    rev1(0, nil)
2.57/1.31	 ||    rev1(s(x0), nil)
2.57/1.31	 ||    rev1(x0, cons(x1, x2))
2.57/1.31	 ||    rev2(x0, nil)
2.57/1.31	 ||    rev2(x0, cons(x1, x2))
2.57/1.31	 || 
2.57/1.31	 || We have to consider all minimal (P,Q,R)-chains.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (17) QDPOrderProof (EQUIVALENT)
2.57/1.31	 || We use the reduction pair processor [LPAR04,JAR06].
2.57/1.31	 || 
2.57/1.31	 || 
2.57/1.31	 || The following pairs can be oriented strictly and are deleted.
2.57/1.31	 || 
2.57/1.31	 ||    REV2(%X, cons(%Y, %Z)) -> REV(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 ||    REV2(%X, cons(%Y, %Z)) -> REV2(%Y, %Z)
2.57/1.31	 || The remaining pairs can at least be oriented weakly.
2.57/1.31	 || Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
2.57/1.31	 || 
2.57/1.31	 || POL( REV_1(x_1) ) = max{0, x_1 - 2}
2.57/1.31	 || POL( cons_2(x_1, x_2) ) = x_2 + 2
2.57/1.31	 || POL( rev2_2(x_1, x_2) ) = x_2
2.57/1.31	 || POL( nil ) = 0
2.57/1.31	 || POL( rev_1(x_1) ) = x_1
2.57/1.31	 || POL( rev1_2(x_1, x_2) ) = x_2
2.57/1.31	 || POL( 0 ) = 2
2.57/1.31	 || POL( s_1(x_1) ) = x_1 + 1
2.57/1.31	 || POL( REV2_2(x_1, x_2) ) = x_2
2.57/1.31	 || 
2.57/1.31	 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
2.57/1.31	 || 
2.57/1.31	 ||    rev2(%X, nil) -> nil
2.57/1.31	 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
2.57/1.31	 || 
2.57/1.31	 || 
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (18)
2.57/1.31	 || Obligation:
2.57/1.31	 || Q DP problem:
2.57/1.31	 || The TRS P consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    REV(cons(%X, %Y)) -> REV2(%X, %Y)
2.57/1.31	 || 
2.57/1.31	 || The TRS R consists of the following rules:
2.57/1.31	 || 
2.57/1.31	 ||    rev2(%X, nil) -> nil
2.57/1.31	 ||    rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z)))
2.57/1.31	 ||    rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y))
2.57/1.31	 ||    rev1(0, nil) -> 0
2.57/1.31	 ||    rev1(s(%X), nil) -> s(%X)
2.57/1.31	 ||    rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z)
2.57/1.31	 || 
2.57/1.31	 || The set Q consists of the following terms:
2.57/1.31	 || 
2.57/1.31	 ||    rev(nil)
2.57/1.31	 ||    rev(cons(x0, x1))
2.57/1.31	 ||    rev1(0, nil)
2.57/1.31	 ||    rev1(s(x0), nil)
2.57/1.31	 ||    rev1(x0, cons(x1, x2))
2.57/1.31	 ||    rev2(x0, nil)
2.57/1.31	 ||    rev2(x0, cons(x1, x2))
2.57/1.31	 || 
2.57/1.31	 || We have to consider all minimal (P,Q,R)-chains.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (19) DependencyGraphProof (EQUIVALENT)
2.57/1.31	 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
2.57/1.31	 || ----------------------------------------
2.57/1.31	 || 
2.57/1.31	 || (20)
2.57/1.31	 || TRUE
2.57/1.31	 || 
2.57/1.31	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]).
2.57/1.31	
2.57/1.31	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
2.57/1.31	
2.57/1.31	  Dependency Pairs P_0:
2.57/1.31	
2.57/1.31	    0] map#(F, cons(X, Y)) =#> map#(F, Y)   
2.57/1.31	    1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
2.57/1.31	    2] filter2#(true, F, X, Y) =#> filter#(F, Y)   
2.57/1.31	    3] filter2#(false, F, X, Y) =#> filter#(F, Y)   
2.57/1.31	
2.57/1.31	  Rules R_0:
2.57/1.31	
2.57/1.31	    rev(nil) => nil 
2.57/1.31	    rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) 
2.57/1.31	    rev1(0, nil) => 0 
2.57/1.31	    rev1(s(X), nil) => s(X) 
2.57/1.31	    rev1(X, cons(Y, Z)) => rev1(Y, Z) 
2.57/1.31	    rev2(X, nil) => nil 
2.57/1.31	    rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) 
2.57/1.31	    map(F, nil) => nil 
2.57/1.31	    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
2.57/1.31	    filter(F, nil) => nil 
2.57/1.31	    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
2.57/1.31	    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
2.57/1.31	    filter2(false, F, X, Y) => filter(F, Y) 
2.57/1.31	
2.57/1.31	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
2.57/1.31	
2.57/1.31	We consider the dependency pair problem (P_0, R_0, static, formative).
2.57/1.31	
2.57/1.31	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
2.57/1.31	
2.57/1.31	    * 0 : 0 
2.57/1.31	    * 1 : 2, 3 
2.57/1.31	    * 2 : 1 
2.57/1.31	    * 3 : 1 
2.57/1.31	
2.57/1.31	This graph has the following strongly connected components:
2.57/1.31	
2.57/1.31	  P_1:
2.57/1.31	
2.57/1.31	    map#(F, cons(X, Y)) =#> map#(F, Y)   
2.57/1.31	
2.57/1.31	  P_2:
2.57/1.31	
2.57/1.31	    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
2.57/1.31	    filter2#(true, F, X, Y) =#> filter#(F, Y)   
2.57/1.31	    filter2#(false, F, X, Y) =#> filter#(F, Y)   
2.57/1.31	
2.57/1.31	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).
2.57/1.31	
2.57/1.31	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.
2.57/1.31	
2.57/1.31	We consider the dependency pair problem (P_2, R_0, static, formative).
2.57/1.31	
2.57/1.31	We apply the subterm criterion with the following projection function:
2.57/1.31	
2.57/1.31	  nu(filter2#) = 4 
2.57/1.31	  nu(filter#) = 2 
2.57/1.31	
2.57/1.31	Thus, we can orient the dependency pairs as follows:
2.57/1.31	
2.57/1.31	  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
2.57/1.31	  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
2.57/1.31	  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
2.57/1.31	
2.57/1.31	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:
2.57/1.31	
2.57/1.31	  filter2#(true, F, X, Y) =#> filter#(F, Y)   
2.57/1.31	  filter2#(false, F, X, Y) =#> filter#(F, Y)   
2.57/1.31	
2.57/1.31	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite.
2.57/1.31	
2.57/1.31	We consider the dependency pair problem (P_3, R_0, static, formative).
2.57/1.31	
2.57/1.31	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
2.57/1.31	
2.57/1.31	    * 0 :  
2.57/1.31	    * 1 :  
2.57/1.31	
2.57/1.31	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
2.57/1.31	
2.57/1.31	Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.
2.57/1.31	
2.57/1.31	We consider the dependency pair problem (P_1, R_0, static, formative).
2.57/1.31	
2.57/1.31	We apply the subterm criterion with the following projection function:
2.57/1.31	
2.57/1.31	  nu(map#) = 2 
2.57/1.31	
2.57/1.31	Thus, we can orient the dependency pairs as follows:
2.57/1.31	
2.57/1.31	  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 
2.57/1.31	
2.57/1.31	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.
2.57/1.31	
2.57/1.31	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
2.57/1.31	
2.57/1.31	
2.57/1.31	+++ Citations +++
2.57/1.31	
2.57/1.31	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
2.57/1.31	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
2.57/1.31	[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.
2.57/1.31	EOF