/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 21 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (29) YES
    (30) QDP
        (31) UsableRulesProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (34) YES
    (35) QDP
        (36) UsableRulesProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (39) YES
    (40) QDP
        (41) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (42) YES
    (43) QDP
        (44) UsableRulesProof [EQUIVALENT, 0 ms]
        (45) QDP
        (46) MRRProof [EQUIVALENT, 2 ms]
        (47) QDP
        (48) PisEmptyProof [EQUIVALENT, 0 ms]
        (49) YES
    (50) QDP
        (51) UsableRulesProof [EQUIVALENT, 0 ms]
        (52) QDP
        (53) MRRProof [EQUIVALENT, 3 ms]
        (54) QDP
        (55) PisEmptyProof [EQUIVALENT, 0 ms]
        (56) YES
    (57) QDP
        (58) UsableRulesProof [EQUIVALENT, 0 ms]
        (59) QDP
        (60) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (61) YES
    (62) QDP
        (63) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (64) YES
    (65) QDP
        (66) UsableRulesProof [EQUIVALENT, 0 ms]
        (67) QDP
        (68) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (69) YES
    (70) QDP
        (71) UsableRulesProof [EQUIVALENT, 0 ms]
        (72) QDP
        (73) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (74) YES


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

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.

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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   +^1(0(x), 0(y)) -> 0^1(+(x, y))
   +^1(0(x), 0(y)) -> +^1(x, y)
   +^1(0(x), 1(y)) -> +^1(x, y)
   +^1(1(x), 0(y)) -> +^1(x, y)
   +^1(1(x), 1(y)) -> 0^1(+(+(x, y), 1(#)))
   +^1(1(x), 1(y)) -> +^1(+(x, y), 1(#))
   +^1(1(x), 1(y)) -> +^1(x, y)
   +^1(+(x, y), z) -> +^1(x, +(y, z))
   +^1(+(x, y), z) -> +^1(y, z)
   -^1(0(x), 0(y)) -> 0^1(-(x, y))
   -^1(0(x), 0(y)) -> -^1(x, y)
   -^1(0(x), 1(y)) -> -^1(-(x, y), 1(#))
   -^1(0(x), 1(y)) -> -^1(x, y)
   -^1(1(x), 0(y)) -> -^1(x, y)
   -^1(1(x), 1(y)) -> 0^1(-(x, y))
   -^1(1(x), 1(y)) -> -^1(x, y)
   EQ(#, 0(y)) -> EQ(#, y)
   EQ(0(x), #) -> EQ(x, #)
   EQ(1(x), 1(y)) -> EQ(x, y)
   EQ(0(x), 0(y)) -> EQ(x, y)
   GE(0(x), 0(y)) -> GE(x, y)
   GE(0(x), 1(y)) -> NOT(ge(y, x))
   GE(0(x), 1(y)) -> GE(y, x)
   GE(1(x), 0(y)) -> GE(x, y)
   GE(1(x), 1(y)) -> GE(x, y)
   GE(#, 0(x)) -> GE(#, x)
   LOG(x) -> -^1(log'(x), 1(#))
   LOG(x) -> LOG'(x)
   LOG'(1(x)) -> +^1(log'(x), 1(#))
   LOG'(1(x)) -> LOG'(x)
   LOG'(0(x)) -> IF(ge(x, 1(#)), +(log'(x), 1(#)), #)
   LOG'(0(x)) -> GE(x, 1(#))
   LOG'(0(x)) -> +^1(log'(x), 1(#))
   LOG'(0(x)) -> LOG'(x)
   *^1(0(x), y) -> 0^1(*(x, y))
   *^1(0(x), y) -> *^1(x, y)
   *^1(1(x), y) -> +^1(0(*(x, y)), y)
   *^1(1(x), y) -> 0^1(*(x, y))
   *^1(1(x), y) -> *^1(x, y)
   *^1(*(x, y), z) -> *^1(x, *(y, z))
   *^1(*(x, y), z) -> *^1(y, z)
   *^1(x, +(y, z)) -> +^1(*(x, y), *(x, z))
   *^1(x, +(y, z)) -> *^1(x, y)
   *^1(x, +(y, z)) -> *^1(x, z)
   APP(cons(x, l1), l2) -> APP(l1, l2)
   SUM(nil) -> 0^1(#)
   SUM(cons(x, l)) -> +^1(x, sum(l))
   SUM(cons(x, l)) -> SUM(l)
   SUM(app(l1, l2)) -> +^1(sum(l1), sum(l2))
   SUM(app(l1, l2)) -> SUM(l1)
   SUM(app(l1, l2)) -> SUM(l2)
   PROD(cons(x, l)) -> *^1(x, prod(l))
   PROD(cons(x, l)) -> PROD(l)
   PROD(app(l1, l2)) -> *^1(prod(l1), prod(l2))
   PROD(app(l1, l2)) -> PROD(l1)
   PROD(app(l1, l2)) -> PROD(l2)
   MEM(x, cons(y, l)) -> IF(eq(x, y), true, mem(x, l))
   MEM(x, cons(y, l)) -> EQ(x, y)
   MEM(x, cons(y, l)) -> MEM(x, l)
   INTER(app(l1, l2), l3) -> APP(inter(l1, l3), inter(l2, l3))
   INTER(app(l1, l2), l3) -> INTER(l1, l3)
   INTER(app(l1, l2), l3) -> INTER(l2, l3)
   INTER(l1, app(l2, l3)) -> APP(inter(l1, l2), inter(l1, l3))
   INTER(l1, app(l2, l3)) -> INTER(l1, l2)
   INTER(l1, app(l2, l3)) -> INTER(l1, l3)
   INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2)
   INTER(cons(x, l1), l2) -> MEM(x, l2)
   INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1)
   INTER(l1, cons(x, l2)) -> MEM(x, l1)
   IFINTER(true, x, l1, l2) -> INTER(l1, l2)
   IFINTER(false, x, l1, l2) -> INTER(l1, l2)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 14 SCCs with 26 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(cons(x, l1), l2) -> APP(l1, l2)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(cons(x, l1), l2) -> APP(l1, l2)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(8) 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:
*APP(cons(x, l1), l2) -> APP(l1, l2)
The graph contains the following edges 1 > 1, 2 >= 2


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

(9)
YES

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

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(#, 0(x)) -> GE(#, x)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(#, 0(x)) -> GE(#, x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(13) 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:
*GE(#, 0(x)) -> GE(#, x)
The graph contains the following edges 1 >= 1, 2 > 2


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

(14)
YES

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

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(0(x), 1(y)) -> GE(y, x)
   GE(0(x), 0(y)) -> GE(x, y)
   GE(1(x), 0(y)) -> GE(x, y)
   GE(1(x), 1(y)) -> GE(x, y)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(16) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(0(x), 1(y)) -> GE(y, x)
   GE(0(x), 0(y)) -> GE(x, y)
   GE(1(x), 0(y)) -> GE(x, y)
   GE(1(x), 1(y)) -> GE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(18) 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:
*GE(0(x), 1(y)) -> GE(y, x)
The graph contains the following edges 2 > 1, 1 > 2


*GE(0(x), 0(y)) -> GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2


*GE(1(x), 0(y)) -> GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2


*GE(1(x), 1(y)) -> GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(19)
YES

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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(0(x), #) -> EQ(x, #)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(21) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(0(x), #) -> EQ(x, #)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(23) 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(0(x), #) -> EQ(x, #)
The graph contains the following edges 1 > 1, 2 >= 2


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

(24)
YES

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

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(#, 0(y)) -> EQ(#, y)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(26) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(#, 0(y)) -> EQ(#, 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(#, 0(y)) -> EQ(#, 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:

   EQ(0(x), 0(y)) -> EQ(x, y)
   EQ(1(x), 1(y)) -> EQ(x, y)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(31) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(0(x), 0(y)) -> EQ(x, y)
   EQ(1(x), 1(y)) -> EQ(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(33) 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(0(x), 0(y)) -> EQ(x, y)
The graph contains the following edges 1 > 1, 2 > 2


*EQ(1(x), 1(y)) -> EQ(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(34)
YES

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

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MEM(x, cons(y, l)) -> MEM(x, l)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(36) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MEM(x, cons(y, l)) -> MEM(x, l)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(38) 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:
*MEM(x, cons(y, l)) -> MEM(x, l)
The graph contains the following edges 1 >= 1, 2 > 2


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

(39)
YES

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

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INTER(app(l1, l2), l3) -> INTER(l2, l3)
   INTER(app(l1, l2), l3) -> INTER(l1, l3)
   INTER(l1, app(l2, l3)) -> INTER(l1, l2)
   INTER(l1, app(l2, l3)) -> INTER(l1, l3)
   INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2)
   IFINTER(true, x, l1, l2) -> INTER(l1, l2)
   INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1)
   IFINTER(false, x, l1, l2) -> INTER(l1, l2)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(41) 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:
*INTER(app(l1, l2), l3) -> INTER(l2, l3)
The graph contains the following edges 1 > 1, 2 >= 2


*INTER(app(l1, l2), l3) -> INTER(l1, l3)
The graph contains the following edges 1 > 1, 2 >= 2


*INTER(l1, app(l2, l3)) -> INTER(l1, l2)
The graph contains the following edges 1 >= 1, 2 > 2


*INTER(l1, app(l2, l3)) -> INTER(l1, l3)
The graph contains the following edges 1 >= 1, 2 > 2


*INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2)
The graph contains the following edges 1 > 2, 1 > 3, 2 >= 4


*INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1)
The graph contains the following edges 2 > 2, 2 > 3, 1 >= 4


*IFINTER(true, x, l1, l2) -> INTER(l1, l2)
The graph contains the following edges 3 >= 1, 4 >= 2


*IFINTER(false, x, l1, l2) -> INTER(l1, l2)
The graph contains the following edges 3 >= 1, 4 >= 2


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

(42)
YES

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

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   -^1(0(x), 1(y)) -> -^1(-(x, y), 1(#))
   -^1(0(x), 1(y)) -> -^1(x, y)
   -^1(0(x), 0(y)) -> -^1(x, y)
   -^1(1(x), 0(y)) -> -^1(x, y)
   -^1(1(x), 1(y)) -> -^1(x, y)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(44) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(45)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   -^1(0(x), 1(y)) -> -^1(-(x, y), 1(#))
   -^1(0(x), 1(y)) -> -^1(x, y)
   -^1(0(x), 0(y)) -> -^1(x, y)
   -^1(1(x), 0(y)) -> -^1(x, y)
   -^1(1(x), 1(y)) -> -^1(x, y)

The TRS R consists of the following rules:

   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   0(#) -> #

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(46) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   -^1(0(x), 1(y)) -> -^1(-(x, y), 1(#))
   -^1(0(x), 1(y)) -> -^1(x, y)
   -^1(0(x), 0(y)) -> -^1(x, y)
   -^1(1(x), 0(y)) -> -^1(x, y)
   -^1(1(x), 1(y)) -> -^1(x, y)

Strictly oriented rules of the TRS R:

   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   0(#) -> #

Used ordering: Knuth-Bendix order [KBO] with precedence:-_2 > -^1_2 > 0_1 > 1_1 > #

and weight map:

   #=2
   0_1=5
   1_1=3
   -_2=0
   -^1_2=0

The variable weight is 1

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

(47)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(48) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(49)
YES

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

(50)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   +^1(0(x), 1(y)) -> +^1(x, y)
   +^1(0(x), 0(y)) -> +^1(x, y)
   +^1(1(x), 0(y)) -> +^1(x, y)
   +^1(1(x), 1(y)) -> +^1(+(x, y), 1(#))
   +^1(1(x), 1(y)) -> +^1(x, y)
   +^1(+(x, y), z) -> +^1(x, +(y, z))
   +^1(+(x, y), z) -> +^1(y, z)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(51) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(52)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   +^1(0(x), 1(y)) -> +^1(x, y)
   +^1(0(x), 0(y)) -> +^1(x, y)
   +^1(1(x), 0(y)) -> +^1(x, y)
   +^1(1(x), 1(y)) -> +^1(+(x, y), 1(#))
   +^1(1(x), 1(y)) -> +^1(x, y)
   +^1(+(x, y), z) -> +^1(x, +(y, z))
   +^1(+(x, y), z) -> +^1(y, z)

The TRS R consists of the following rules:

   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   0(#) -> #

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(53) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   +^1(0(x), 1(y)) -> +^1(x, y)
   +^1(0(x), 0(y)) -> +^1(x, y)
   +^1(1(x), 0(y)) -> +^1(x, y)
   +^1(1(x), 1(y)) -> +^1(+(x, y), 1(#))
   +^1(1(x), 1(y)) -> +^1(x, y)
   +^1(+(x, y), z) -> +^1(x, +(y, z))
   +^1(+(x, y), z) -> +^1(y, z)

Strictly oriented rules of the TRS R:

   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   0(#) -> #

Used ordering: Knuth-Bendix order [KBO] with precedence:# > +_2 > 1_1 > 0_1 > +^1_2

and weight map:

   #=2
   0_1=1
   1_1=3
   +_2=0
   +^1_2=0

The variable weight is 1

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

(54)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(55) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(56)
YES

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

(57)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(app(l1, l2)) -> SUM(l1)
   SUM(cons(x, l)) -> SUM(l)
   SUM(app(l1, l2)) -> SUM(l2)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(58) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(59)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(app(l1, l2)) -> SUM(l1)
   SUM(cons(x, l)) -> SUM(l)
   SUM(app(l1, l2)) -> SUM(l2)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(60) 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:
*SUM(app(l1, l2)) -> SUM(l1)
The graph contains the following edges 1 > 1


*SUM(cons(x, l)) -> SUM(l)
The graph contains the following edges 1 > 1


*SUM(app(l1, l2)) -> SUM(l2)
The graph contains the following edges 1 > 1


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

(61)
YES

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

(62)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   *^1(1(x), y) -> *^1(x, y)
   *^1(0(x), y) -> *^1(x, y)
   *^1(*(x, y), z) -> *^1(x, *(y, z))
   *^1(*(x, y), z) -> *^1(y, z)
   *^1(x, +(y, z)) -> *^1(x, y)
   *^1(x, +(y, z)) -> *^1(x, z)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(63) 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:
**^1(1(x), y) -> *^1(x, y)
The graph contains the following edges 1 > 1, 2 >= 2


**^1(0(x), y) -> *^1(x, y)
The graph contains the following edges 1 > 1, 2 >= 2


**^1(*(x, y), z) -> *^1(x, *(y, z))
The graph contains the following edges 1 > 1


**^1(*(x, y), z) -> *^1(y, z)
The graph contains the following edges 1 > 1, 2 >= 2


**^1(x, +(y, z)) -> *^1(x, y)
The graph contains the following edges 1 >= 1, 2 > 2


**^1(x, +(y, z)) -> *^1(x, z)
The graph contains the following edges 1 >= 1, 2 > 2


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

(64)
YES

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

(65)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PROD(app(l1, l2)) -> PROD(l1)
   PROD(cons(x, l)) -> PROD(l)
   PROD(app(l1, l2)) -> PROD(l2)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(66) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(67)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PROD(app(l1, l2)) -> PROD(l1)
   PROD(cons(x, l)) -> PROD(l)
   PROD(app(l1, l2)) -> PROD(l2)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(68) 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:
*PROD(app(l1, l2)) -> PROD(l1)
The graph contains the following edges 1 > 1


*PROD(cons(x, l)) -> PROD(l)
The graph contains the following edges 1 > 1


*PROD(app(l1, l2)) -> PROD(l2)
The graph contains the following edges 1 > 1


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

(69)
YES

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

(70)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LOG'(0(x)) -> LOG'(x)
   LOG'(1(x)) -> LOG'(x)

The TRS R consists of the following rules:

   0(#) -> #
   +(x, #) -> x
   +(#, x) -> x
   +(0(x), 0(y)) -> 0(+(x, y))
   +(0(x), 1(y)) -> 1(+(x, y))
   +(1(x), 0(y)) -> 1(+(x, y))
   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
   +(+(x, y), z) -> +(x, +(y, z))
   -(#, x) -> #
   -(x, #) -> x
   -(0(x), 0(y)) -> 0(-(x, y))
   -(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
   -(1(x), 0(y)) -> 1(-(x, y))
   -(1(x), 1(y)) -> 0(-(x, y))
   not(true) -> false
   not(false) -> true
   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(#, #) -> true
   eq(#, 1(y)) -> false
   eq(1(x), #) -> false
   eq(#, 0(y)) -> eq(#, y)
   eq(0(x), #) -> eq(x, #)
   eq(1(x), 1(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(0(x), 0(y)) -> eq(x, y)
   ge(0(x), 0(y)) -> ge(x, y)
   ge(0(x), 1(y)) -> not(ge(y, x))
   ge(1(x), 0(y)) -> ge(x, y)
   ge(1(x), 1(y)) -> ge(x, y)
   ge(x, #) -> true
   ge(#, 0(x)) -> ge(#, x)
   ge(#, 1(x)) -> false
   log(x) -> -(log'(x), 1(#))
   log'(#) -> #
   log'(1(x)) -> +(log'(x), 1(#))
   log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
   *(#, x) -> #
   *(0(x), y) -> 0(*(x, y))
   *(1(x), y) -> +(0(*(x, y)), y)
   *(*(x, y), z) -> *(x, *(y, z))
   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   sum(nil) -> 0(#)
   sum(cons(x, l)) -> +(x, sum(l))
   sum(app(l1, l2)) -> +(sum(l1), sum(l2))
   prod(nil) -> 1(#)
   prod(cons(x, l)) -> *(x, prod(l))
   prod(app(l1, l2)) -> *(prod(l1), prod(l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
   inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
   inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
   inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
   ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
   ifinter(false, x, l1, l2) -> inter(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(71) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(72)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LOG'(0(x)) -> LOG'(x)
   LOG'(1(x)) -> LOG'(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(73) 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:
*LOG'(0(x)) -> LOG'(x)
The graph contains the following edges 1 > 1


*LOG'(1(x)) -> LOG'(x)
The graph contains the following edges 1 > 1


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

(74)
YES