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


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WORST_CASE(Omega(n^1), O(n^2))
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2).

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 2 ms]
(2) CdtProblem
    (3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
    (4) CdtProblem
    (5) CdtRhsSimplificationProcessorProof [BOTH BOUNDS(ID, ID), 0 ms]
    (6) CdtProblem
    (7) CdtGraphSplitRhsProof [BOTH BOUNDS(ID, ID), 0 ms]
    (8) CdtProblem
    (9) CdtLeafRemovalProof [ComplexityIfPolyImplication, 0 ms]
    (10) CdtProblem
    (11) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
    (12) CdtProblem
    (13) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 3 ms]
    (14) CdtProblem
    (15) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 86 ms]
    (16) CdtProblem
    (17) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 55 ms]
    (18) CdtProblem
    (19) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 38 ms]
    (20) CdtProblem
    (21) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 28 ms]
    (22) CdtProblem
    (23) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
    (24) BOUNDS(1, 1)
(25) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(26) TRS for Loop Detection
    (27) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
    (28) BEST
        (29) proven lower bound
            (30) LowerBoundPropagationProof [FINISHED, 0 ms]
            (31) BOUNDS(n^1, INF)
        (32) TRS for Loop Detection


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

(0)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2).


The TRS R consists of the following rules:

   if(true, t, e) -> t
   if(false, t, e) -> e
   member(x, nil) -> false
   member(x, cons(y, ys)) -> if(eq(x, y), true, member(x, ys))
   eq(nil, nil) -> true
   eq(O(x), 0(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(1(x), 1(y)) -> eq(x, y)
   negate(0(x)) -> 1(x)
   negate(1(x)) -> 0(x)
   choice(cons(x, xs)) -> x
   choice(cons(x, xs)) -> choice(xs)
   guess(nil) -> nil
   guess(cons(clause, cnf)) -> cons(choice(clause), guess(cnf))
   verify(nil) -> true
   verify(cons(l, ls)) -> if(member(negate(l), ls), false, verify(ls))
   sat(cnf) -> satck(cnf, guess(cnf))
   satck(cnf, assign) -> if(verify(assign), assign, unsat)

S is empty.
Rewrite Strategy: INNERMOST
----------------------------------------

(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS to CDT
----------------------------------------

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   if(true, z0, z1) -> z0
   if(false, z0, z1) -> z1
   member(z0, nil) -> false
   member(z0, cons(z1, z2)) -> if(eq(z0, z1), true, member(z0, z2))
   eq(nil, nil) -> true
   eq(O(z0), 0(z1)) -> eq(z0, z1)
   eq(0(z0), 1(z1)) -> false
   eq(1(z0), 0(z1)) -> false
   eq(1(z0), 1(z1)) -> eq(z0, z1)
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   verify(nil) -> true
   verify(cons(z0, z1)) -> if(member(negate(z0), z1), false, verify(z1))
   sat(z0) -> satck(z0, guess(z0))
   satck(z0, z1) -> if(verify(z1), z1, unsat)
Tuples:
   IF(true, z0, z1) -> c
   IF(false, z0, z1) -> c1
   MEMBER(z0, nil) -> c2
   MEMBER(z0, cons(z1, z2)) -> c3(IF(eq(z0, z1), true, member(z0, z2)), EQ(z0, z1), MEMBER(z0, z2))
   EQ(nil, nil) -> c4
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(0(z0), 1(z1)) -> c6
   EQ(1(z0), 0(z1)) -> c7
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   NEGATE(0(z0)) -> c9
   NEGATE(1(z0)) -> c10
   CHOICE(cons(z0, z1)) -> c11
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(nil) -> c13
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   VERIFY(nil) -> c15
   VERIFY(cons(z0, z1)) -> c16(IF(member(negate(z0), z1), false, verify(z1)), MEMBER(negate(z0), z1), NEGATE(z0), VERIFY(z1))
   SAT(z0) -> c17(SATCK(z0, guess(z0)), GUESS(z0))
   SATCK(z0, z1) -> c18(IF(verify(z1), z1, unsat), VERIFY(z1))
S tuples:
   IF(true, z0, z1) -> c
   IF(false, z0, z1) -> c1
   MEMBER(z0, nil) -> c2
   MEMBER(z0, cons(z1, z2)) -> c3(IF(eq(z0, z1), true, member(z0, z2)), EQ(z0, z1), MEMBER(z0, z2))
   EQ(nil, nil) -> c4
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(0(z0), 1(z1)) -> c6
   EQ(1(z0), 0(z1)) -> c7
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   NEGATE(0(z0)) -> c9
   NEGATE(1(z0)) -> c10
   CHOICE(cons(z0, z1)) -> c11
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(nil) -> c13
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   VERIFY(nil) -> c15
   VERIFY(cons(z0, z1)) -> c16(IF(member(negate(z0), z1), false, verify(z1)), MEMBER(negate(z0), z1), NEGATE(z0), VERIFY(z1))
   SAT(z0) -> c17(SATCK(z0, guess(z0)), GUESS(z0))
   SATCK(z0, z1) -> c18(IF(verify(z1), z1, unsat), VERIFY(z1))
K tuples:none
Defined Rule Symbols:   if_3, member_2, eq_2, negate_1, choice_1, guess_1, verify_1, sat_1, satck_2

Defined Pair Symbols:   IF_3, MEMBER_2, EQ_2, NEGATE_1, CHOICE_1, GUESS_1, VERIFY_1, SAT_1, SATCK_2

Compound Symbols:   c, c1, c2, c3_3, c4, c5_1, c6, c7, c8_1, c9, c10, c11, c12_1, c13, c14_2, c15, c16_4, c17_2, c18_2


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

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 11 trailing nodes:
   IF(true, z0, z1) -> c
   NEGATE(0(z0)) -> c9
   GUESS(nil) -> c13
   CHOICE(cons(z0, z1)) -> c11
   IF(false, z0, z1) -> c1
   EQ(nil, nil) -> c4
   EQ(1(z0), 0(z1)) -> c7
   NEGATE(1(z0)) -> c10
   EQ(0(z0), 1(z1)) -> c6
   MEMBER(z0, nil) -> c2
   VERIFY(nil) -> c15

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

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   if(true, z0, z1) -> z0
   if(false, z0, z1) -> z1
   member(z0, nil) -> false
   member(z0, cons(z1, z2)) -> if(eq(z0, z1), true, member(z0, z2))
   eq(nil, nil) -> true
   eq(O(z0), 0(z1)) -> eq(z0, z1)
   eq(0(z0), 1(z1)) -> false
   eq(1(z0), 0(z1)) -> false
   eq(1(z0), 1(z1)) -> eq(z0, z1)
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   verify(nil) -> true
   verify(cons(z0, z1)) -> if(member(negate(z0), z1), false, verify(z1))
   sat(z0) -> satck(z0, guess(z0))
   satck(z0, z1) -> if(verify(z1), z1, unsat)
Tuples:
   MEMBER(z0, cons(z1, z2)) -> c3(IF(eq(z0, z1), true, member(z0, z2)), EQ(z0, z1), MEMBER(z0, z2))
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   VERIFY(cons(z0, z1)) -> c16(IF(member(negate(z0), z1), false, verify(z1)), MEMBER(negate(z0), z1), NEGATE(z0), VERIFY(z1))
   SAT(z0) -> c17(SATCK(z0, guess(z0)), GUESS(z0))
   SATCK(z0, z1) -> c18(IF(verify(z1), z1, unsat), VERIFY(z1))
S tuples:
   MEMBER(z0, cons(z1, z2)) -> c3(IF(eq(z0, z1), true, member(z0, z2)), EQ(z0, z1), MEMBER(z0, z2))
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   VERIFY(cons(z0, z1)) -> c16(IF(member(negate(z0), z1), false, verify(z1)), MEMBER(negate(z0), z1), NEGATE(z0), VERIFY(z1))
   SAT(z0) -> c17(SATCK(z0, guess(z0)), GUESS(z0))
   SATCK(z0, z1) -> c18(IF(verify(z1), z1, unsat), VERIFY(z1))
K tuples:none
Defined Rule Symbols:   if_3, member_2, eq_2, negate_1, choice_1, guess_1, verify_1, sat_1, satck_2

Defined Pair Symbols:   MEMBER_2, EQ_2, CHOICE_1, GUESS_1, VERIFY_1, SAT_1, SATCK_2

Compound Symbols:   c3_3, c5_1, c8_1, c12_1, c14_2, c16_4, c17_2, c18_2


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

(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
Removed 4 trailing tuple parts
----------------------------------------

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   if(true, z0, z1) -> z0
   if(false, z0, z1) -> z1
   member(z0, nil) -> false
   member(z0, cons(z1, z2)) -> if(eq(z0, z1), true, member(z0, z2))
   eq(nil, nil) -> true
   eq(O(z0), 0(z1)) -> eq(z0, z1)
   eq(0(z0), 1(z1)) -> false
   eq(1(z0), 0(z1)) -> false
   eq(1(z0), 1(z1)) -> eq(z0, z1)
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   verify(nil) -> true
   verify(cons(z0, z1)) -> if(member(negate(z0), z1), false, verify(z1))
   sat(z0) -> satck(z0, guess(z0))
   satck(z0, z1) -> if(verify(z1), z1, unsat)
Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   SAT(z0) -> c17(SATCK(z0, guess(z0)), GUESS(z0))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
S tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   SAT(z0) -> c17(SATCK(z0, guess(z0)), GUESS(z0))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
K tuples:none
Defined Rule Symbols:   if_3, member_2, eq_2, negate_1, choice_1, guess_1, verify_1, sat_1, satck_2

Defined Pair Symbols:   EQ_2, CHOICE_1, GUESS_1, SAT_1, MEMBER_2, VERIFY_1, SATCK_2

Compound Symbols:   c5_1, c8_1, c12_1, c14_2, c17_2, c3_2, c16_2, c18_1


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

(7) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID))
Split RHS of tuples not part of any SCC
----------------------------------------

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   if(true, z0, z1) -> z0
   if(false, z0, z1) -> z1
   member(z0, nil) -> false
   member(z0, cons(z1, z2)) -> if(eq(z0, z1), true, member(z0, z2))
   eq(nil, nil) -> true
   eq(O(z0), 0(z1)) -> eq(z0, z1)
   eq(0(z0), 1(z1)) -> false
   eq(1(z0), 0(z1)) -> false
   eq(1(z0), 1(z1)) -> eq(z0, z1)
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   verify(nil) -> true
   verify(cons(z0, z1)) -> if(member(negate(z0), z1), false, verify(z1))
   sat(z0) -> satck(z0, guess(z0))
   satck(z0, z1) -> if(verify(z1), z1, unsat)
Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
   SAT(z0) -> c(GUESS(z0))
S tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
   SAT(z0) -> c(GUESS(z0))
K tuples:none
Defined Rule Symbols:   if_3, member_2, eq_2, negate_1, choice_1, guess_1, verify_1, sat_1, satck_2

Defined Pair Symbols:   EQ_2, CHOICE_1, GUESS_1, MEMBER_2, VERIFY_1, SATCK_2, SAT_1

Compound Symbols:   c5_1, c8_1, c12_1, c14_2, c3_2, c16_2, c18_1, c_1


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

(9) CdtLeafRemovalProof (ComplexityIfPolyImplication)
Removed 1 leading nodes:
   SAT(z0) -> c(GUESS(z0))

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

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   if(true, z0, z1) -> z0
   if(false, z0, z1) -> z1
   member(z0, nil) -> false
   member(z0, cons(z1, z2)) -> if(eq(z0, z1), true, member(z0, z2))
   eq(nil, nil) -> true
   eq(O(z0), 0(z1)) -> eq(z0, z1)
   eq(0(z0), 1(z1)) -> false
   eq(1(z0), 0(z1)) -> false
   eq(1(z0), 1(z1)) -> eq(z0, z1)
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   verify(nil) -> true
   verify(cons(z0, z1)) -> if(member(negate(z0), z1), false, verify(z1))
   sat(z0) -> satck(z0, guess(z0))
   satck(z0, z1) -> if(verify(z1), z1, unsat)
Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
S tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
K tuples:none
Defined Rule Symbols:   if_3, member_2, eq_2, negate_1, choice_1, guess_1, verify_1, sat_1, satck_2

Defined Pair Symbols:   EQ_2, CHOICE_1, GUESS_1, MEMBER_2, VERIFY_1, SATCK_2, SAT_1

Compound Symbols:   c5_1, c8_1, c12_1, c14_2, c3_2, c16_2, c18_1, c_1


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

(11) CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
The following tuples could be moved from S to K by knowledge propagation:
   SAT(z0) -> c(SATCK(z0, guess(z0)))
   SATCK(z0, z1) -> c18(VERIFY(z1))

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

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   if(true, z0, z1) -> z0
   if(false, z0, z1) -> z1
   member(z0, nil) -> false
   member(z0, cons(z1, z2)) -> if(eq(z0, z1), true, member(z0, z2))
   eq(nil, nil) -> true
   eq(O(z0), 0(z1)) -> eq(z0, z1)
   eq(0(z0), 1(z1)) -> false
   eq(1(z0), 0(z1)) -> false
   eq(1(z0), 1(z1)) -> eq(z0, z1)
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   verify(nil) -> true
   verify(cons(z0, z1)) -> if(member(negate(z0), z1), false, verify(z1))
   sat(z0) -> satck(z0, guess(z0))
   satck(z0, z1) -> if(verify(z1), z1, unsat)
Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
S tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
K tuples:
   SAT(z0) -> c(SATCK(z0, guess(z0)))
   SATCK(z0, z1) -> c18(VERIFY(z1))
Defined Rule Symbols:   if_3, member_2, eq_2, negate_1, choice_1, guess_1, verify_1, sat_1, satck_2

Defined Pair Symbols:   EQ_2, CHOICE_1, GUESS_1, MEMBER_2, VERIFY_1, SATCK_2, SAT_1

Compound Symbols:   c5_1, c8_1, c12_1, c14_2, c3_2, c16_2, c18_1, c_1


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

(13) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   if(true, z0, z1) -> z0
   if(false, z0, z1) -> z1
   member(z0, nil) -> false
   member(z0, cons(z1, z2)) -> if(eq(z0, z1), true, member(z0, z2))
   eq(nil, nil) -> true
   eq(O(z0), 0(z1)) -> eq(z0, z1)
   eq(0(z0), 1(z1)) -> false
   eq(1(z0), 0(z1)) -> false
   eq(1(z0), 1(z1)) -> eq(z0, z1)
   verify(nil) -> true
   verify(cons(z0, z1)) -> if(member(negate(z0), z1), false, verify(z1))
   sat(z0) -> satck(z0, guess(z0))
   satck(z0, z1) -> if(verify(z1), z1, unsat)

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

(14)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
S tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
K tuples:
   SAT(z0) -> c(SATCK(z0, guess(z0)))
   SATCK(z0, z1) -> c18(VERIFY(z1))
Defined Rule Symbols:   negate_1, guess_1, choice_1

Defined Pair Symbols:   EQ_2, CHOICE_1, GUESS_1, MEMBER_2, VERIFY_1, SATCK_2, SAT_1

Compound Symbols:   c5_1, c8_1, c12_1, c14_2, c3_2, c16_2, c18_1, c_1


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

(15) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
We considered the (Usable) Rules:
   negate(1(z0)) -> 0(z0)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   choice(cons(z0, z1)) -> choice(z1)
   choice(cons(z0, z1)) -> z0
   negate(0(z0)) -> 1(z0)
And the Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0(x_1)) = [1] + x_1
   POL(1(x_1)) = [1] + x_1
   POL(CHOICE(x_1)) = x_1
   POL(EQ(x_1, x_2)) = 0
   POL(GUESS(x_1)) = x_1
   POL(MEMBER(x_1, x_2)) = x_1
   POL(O(x_1)) = 0
   POL(SAT(x_1)) = [1] + x_1
   POL(SATCK(x_1, x_2)) = x_2
   POL(VERIFY(x_1)) = x_1
   POL(c(x_1)) = x_1
   POL(c12(x_1)) = x_1
   POL(c14(x_1, x_2)) = x_1 + x_2
   POL(c16(x_1, x_2)) = x_1 + x_2
   POL(c18(x_1)) = x_1
   POL(c3(x_1, x_2)) = x_1 + x_2
   POL(c5(x_1)) = x_1
   POL(c8(x_1)) = x_1
   POL(choice(x_1)) = x_1
   POL(cons(x_1, x_2)) = [1] + x_1 + x_2
   POL(guess(x_1)) = [1] + x_1
   POL(negate(x_1)) = x_1
   POL(nil) = [1]

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

(16)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
S tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
K tuples:
   SAT(z0) -> c(SATCK(z0, guess(z0)))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
Defined Rule Symbols:   negate_1, guess_1, choice_1

Defined Pair Symbols:   EQ_2, CHOICE_1, GUESS_1, MEMBER_2, VERIFY_1, SATCK_2, SAT_1

Compound Symbols:   c5_1, c8_1, c12_1, c14_2, c3_2, c16_2, c18_1, c_1


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

(17) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
We considered the (Usable) Rules:
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
And the Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0(x_1)) = 0
   POL(1(x_1)) = 0
   POL(CHOICE(x_1)) = [1]
   POL(EQ(x_1, x_2)) = 0
   POL(GUESS(x_1)) = [2]x_1 + [2]x_1^2
   POL(MEMBER(x_1, x_2)) = [1] + x_2
   POL(O(x_1)) = 0
   POL(SAT(x_1)) = [2] + x_1 + [2]x_1^2
   POL(SATCK(x_1, x_2)) = x_2 + x_2^2 + x_1^2
   POL(VERIFY(x_1)) = x_1^2
   POL(c(x_1)) = x_1
   POL(c12(x_1)) = x_1
   POL(c14(x_1, x_2)) = x_1 + x_2
   POL(c16(x_1, x_2)) = x_1 + x_2
   POL(c18(x_1)) = x_1
   POL(c3(x_1, x_2)) = x_1 + x_2
   POL(c5(x_1)) = x_1
   POL(c8(x_1)) = x_1
   POL(choice(x_1)) = [1] + x_1 + [2]x_1^2
   POL(cons(x_1, x_2)) = [1] + x_2
   POL(guess(x_1)) = x_1
   POL(negate(x_1)) = 0
   POL(nil) = 0

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

(18)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
S tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
K tuples:
   SAT(z0) -> c(SATCK(z0, guess(z0)))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
Defined Rule Symbols:   negate_1, guess_1, choice_1

Defined Pair Symbols:   EQ_2, CHOICE_1, GUESS_1, MEMBER_2, VERIFY_1, SATCK_2, SAT_1

Compound Symbols:   c5_1, c8_1, c12_1, c14_2, c3_2, c16_2, c18_1, c_1


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

(19) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
We considered the (Usable) Rules:
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   choice(cons(z0, z1)) -> choice(z1)
   choice(cons(z0, z1)) -> z0
And the Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0(x_1)) = [1] + x_1
   POL(1(x_1)) = x_1
   POL(CHOICE(x_1)) = [1] + [2]x_1 + [2]x_1^2
   POL(EQ(x_1, x_2)) = [2] + [2]x_2
   POL(GUESS(x_1)) = [2]x_1 + [2]x_1^2
   POL(MEMBER(x_1, x_2)) = [2]x_2
   POL(O(x_1)) = 0
   POL(SAT(x_1)) = [2] + [2]x_1 + [2]x_1^2
   POL(SATCK(x_1, x_2)) = [2]x_2^2
   POL(VERIFY(x_1)) = x_1^2
   POL(c(x_1)) = x_1
   POL(c12(x_1)) = x_1
   POL(c14(x_1, x_2)) = x_1 + x_2
   POL(c16(x_1, x_2)) = x_1 + x_2
   POL(c18(x_1)) = x_1
   POL(c3(x_1, x_2)) = x_1 + x_2
   POL(c5(x_1)) = x_1
   POL(c8(x_1)) = x_1
   POL(choice(x_1)) = x_1
   POL(cons(x_1, x_2)) = [1] + x_1 + x_2
   POL(guess(x_1)) = x_1
   POL(negate(x_1)) = 0
   POL(nil) = 0

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

(20)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
S tuples:
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
K tuples:
   SAT(z0) -> c(SATCK(z0, guess(z0)))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
Defined Rule Symbols:   negate_1, guess_1, choice_1

Defined Pair Symbols:   EQ_2, CHOICE_1, GUESS_1, MEMBER_2, VERIFY_1, SATCK_2, SAT_1

Compound Symbols:   c5_1, c8_1, c12_1, c14_2, c3_2, c16_2, c18_1, c_1


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

(21) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
We considered the (Usable) Rules:
   negate(1(z0)) -> 0(z0)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   choice(cons(z0, z1)) -> choice(z1)
   choice(cons(z0, z1)) -> z0
   negate(0(z0)) -> 1(z0)
And the Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0(x_1)) = [1] + x_1
   POL(1(x_1)) = [1] + x_1
   POL(CHOICE(x_1)) = [2]x_1 + [2]x_1^2
   POL(EQ(x_1, x_2)) = x_1*x_2
   POL(GUESS(x_1)) = [2]x_1 + [2]x_1^2
   POL(MEMBER(x_1, x_2)) = x_1*x_2
   POL(O(x_1)) = x_1
   POL(SAT(x_1)) = [1] + [2]x_1 + [2]x_1^2
   POL(SATCK(x_1, x_2)) = x_2^2 + x_1^2
   POL(VERIFY(x_1)) = x_1^2
   POL(c(x_1)) = x_1
   POL(c12(x_1)) = x_1
   POL(c14(x_1, x_2)) = x_1 + x_2
   POL(c16(x_1, x_2)) = x_1 + x_2
   POL(c18(x_1)) = x_1
   POL(c3(x_1, x_2)) = x_1 + x_2
   POL(c5(x_1)) = x_1
   POL(c8(x_1)) = x_1
   POL(choice(x_1)) = x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(guess(x_1)) = x_1
   POL(negate(x_1)) = x_1
   POL(nil) = 0

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

(22)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   negate(0(z0)) -> 1(z0)
   negate(1(z0)) -> 0(z0)
   guess(nil) -> nil
   guess(cons(z0, z1)) -> cons(choice(z0), guess(z1))
   choice(cons(z0, z1)) -> z0
   choice(cons(z0, z1)) -> choice(z1)
Tuples:
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   SAT(z0) -> c(SATCK(z0, guess(z0)))
S tuples:none
K tuples:
   SAT(z0) -> c(SATCK(z0, guess(z0)))
   SATCK(z0, z1) -> c18(VERIFY(z1))
   CHOICE(cons(z0, z1)) -> c12(CHOICE(z1))
   GUESS(cons(z0, z1)) -> c14(CHOICE(z0), GUESS(z1))
   VERIFY(cons(z0, z1)) -> c16(MEMBER(negate(z0), z1), VERIFY(z1))
   MEMBER(z0, cons(z1, z2)) -> c3(EQ(z0, z1), MEMBER(z0, z2))
   EQ(O(z0), 0(z1)) -> c5(EQ(z0, z1))
   EQ(1(z0), 1(z1)) -> c8(EQ(z0, z1))
Defined Rule Symbols:   negate_1, guess_1, choice_1

Defined Pair Symbols:   EQ_2, CHOICE_1, GUESS_1, MEMBER_2, VERIFY_1, SATCK_2, SAT_1

Compound Symbols:   c5_1, c8_1, c12_1, c14_2, c3_2, c16_2, c18_1, c_1


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

(23) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
----------------------------------------

(24)
BOUNDS(1, 1)

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

(25) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
Transformed a relative TRS into a decreasing-loop problem.
----------------------------------------

(26)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2).


The TRS R consists of the following rules:

   if(true, t, e) -> t
   if(false, t, e) -> e
   member(x, nil) -> false
   member(x, cons(y, ys)) -> if(eq(x, y), true, member(x, ys))
   eq(nil, nil) -> true
   eq(O(x), 0(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(1(x), 1(y)) -> eq(x, y)
   negate(0(x)) -> 1(x)
   negate(1(x)) -> 0(x)
   choice(cons(x, xs)) -> x
   choice(cons(x, xs)) -> choice(xs)
   guess(nil) -> nil
   guess(cons(clause, cnf)) -> cons(choice(clause), guess(cnf))
   verify(nil) -> true
   verify(cons(l, ls)) -> if(member(negate(l), ls), false, verify(ls))
   sat(cnf) -> satck(cnf, guess(cnf))
   satck(cnf, assign) -> if(verify(assign), assign, unsat)

S is empty.
Rewrite Strategy: INNERMOST
----------------------------------------

(27) DecreasingLoopProof (LOWER BOUND(ID))
The following loop(s) give(s) rise to the lower bound Omega(n^1):

The rewrite sequence

member(x, cons(y, ys)) ->^+ if(eq(x, y), true, member(x, ys))

gives rise to a decreasing loop by considering the right hand sides subterm at position [2].

The pumping substitution is [ys / cons(y, ys)].

The result substitution is [ ].




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(28)
Complex Obligation (BEST)

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(29)
Obligation:
Proved the lower bound n^1 for the following obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2).


The TRS R consists of the following rules:

   if(true, t, e) -> t
   if(false, t, e) -> e
   member(x, nil) -> false
   member(x, cons(y, ys)) -> if(eq(x, y), true, member(x, ys))
   eq(nil, nil) -> true
   eq(O(x), 0(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(1(x), 1(y)) -> eq(x, y)
   negate(0(x)) -> 1(x)
   negate(1(x)) -> 0(x)
   choice(cons(x, xs)) -> x
   choice(cons(x, xs)) -> choice(xs)
   guess(nil) -> nil
   guess(cons(clause, cnf)) -> cons(choice(clause), guess(cnf))
   verify(nil) -> true
   verify(cons(l, ls)) -> if(member(negate(l), ls), false, verify(ls))
   sat(cnf) -> satck(cnf, guess(cnf))
   satck(cnf, assign) -> if(verify(assign), assign, unsat)

S is empty.
Rewrite Strategy: INNERMOST
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(30) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
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(31)
BOUNDS(n^1, INF)

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(32)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2).


The TRS R consists of the following rules:

   if(true, t, e) -> t
   if(false, t, e) -> e
   member(x, nil) -> false
   member(x, cons(y, ys)) -> if(eq(x, y), true, member(x, ys))
   eq(nil, nil) -> true
   eq(O(x), 0(y)) -> eq(x, y)
   eq(0(x), 1(y)) -> false
   eq(1(x), 0(y)) -> false
   eq(1(x), 1(y)) -> eq(x, y)
   negate(0(x)) -> 1(x)
   negate(1(x)) -> 0(x)
   choice(cons(x, xs)) -> x
   choice(cons(x, xs)) -> choice(xs)
   guess(nil) -> nil
   guess(cons(clause, cnf)) -> cons(choice(clause), guess(cnf))
   verify(nil) -> true
   verify(cons(l, ls)) -> if(member(negate(l), ls), false, verify(ls))
   sat(cnf) -> satck(cnf, guess(cnf))
   satck(cnf, assign) -> if(verify(assign), assign, unsat)

S is empty.
Rewrite Strategy: INNERMOST