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


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


WORST_CASE(Omega(n^1), O(n^2))
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) CpxTRS
(1) NestedDefinedSymbolProof [UPPER BOUND(ID), 0 ms]
(2) CpxTRS
    (3) RcToIrcProof [BOTH BOUNDS(ID, ID), 0 ms]
    (4) CpxTRS
    (5) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
    (6) CdtProblem
    (7) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
    (8) CdtProblem
    (9) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
    (10) CdtProblem
    (11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 107 ms]
    (12) CdtProblem
    (13) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
    (14) CdtProblem
    (15) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 94 ms]
    (16) CdtProblem
    (17) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
    (18) CdtProblem
    (19) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 115 ms]
    (20) CdtProblem
    (21) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
    (22) BOUNDS(1, 1)
(23) RenamingProof [BOTH BOUNDS(ID, ID), 0 ms]
(24) CpxTRS
    (25) TypeInferenceProof [BOTH BOUNDS(ID, ID), 0 ms]
    (26) typed CpxTrs
    (27) OrderProof [LOWER BOUND(ID), 0 ms]
    (28) typed CpxTrs
    (29) RewriteLemmaProof [LOWER BOUND(ID), 306 ms]
    (30) BEST
        (31) proven lower bound
            (32) LowerBoundPropagationProof [FINISHED, 0 ms]
            (33) BOUNDS(n^1, INF)
        (34) typed CpxTrs
            (35) RewriteLemmaProof [LOWER BOUND(ID), 77 ms]
            (36) typed CpxTrs


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

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


The TRS R consists of the following rules:

   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(0, 0) -> true
   eq(0, s(x)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   app(app(l1, l2), l3) -> app(l1, app(l2, l3))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
   ifmem(true, x, l) -> true
   ifmem(false, x, l) -> 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)

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

(1) NestedDefinedSymbolProof (UPPER BOUND(ID))
The following defined symbols can occur below the 0th argument of ifinter: ifmem, eq, mem
The following defined symbols can occur below the 0th argument of ifmem: eq

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))

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

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


The TRS R consists of the following rules:

   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(0, 0) -> true
   eq(0, s(x)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
   ifmem(true, x, l) -> true
   ifmem(false, x, l) -> mem(x, l)
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   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)

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

(3) RcToIrcProof (BOTH BOUNDS(ID, ID))
Converted rc-obligation to irc-obligation.

The duplicating contexts are:
mem([], cons(y, l))
inter(cons([], l1), l2)
inter(cons(x, l1), [])
inter(l1, cons([], l2))
inter([], cons(x, l2))


The defined contexts are:
ifinter([], x1, x2, x3)
ifmem([], x1, x2)


[] just represents basic- or constructor-terms in the following defined contexts:
ifmem([], x1, x2)


As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.
----------------------------------------

(4)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).


The TRS R consists of the following rules:

   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(0, 0) -> true
   eq(0, s(x)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
   ifmem(true, x, l) -> true
   ifmem(false, x, l) -> mem(x, l)
   inter(x, nil) -> nil
   inter(nil, x) -> nil
   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)

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

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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   if(true, z0, z1) -> z0
   if(false, z0, z1) -> z1
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   app(nil, z0) -> z0
   app(cons(z0, z1), z2) -> cons(z0, app(z1, z2))
   mem(z0, nil) -> false
   mem(z0, cons(z1, z2)) -> ifmem(eq(z0, z1), z0, z2)
   ifmem(true, z0, z1) -> true
   ifmem(false, z0, z1) -> mem(z0, z1)
   inter(z0, nil) -> nil
   inter(nil, z0) -> nil
   inter(cons(z0, z1), z2) -> ifinter(mem(z0, z2), z0, z1, z2)
   inter(z0, cons(z1, z2)) -> ifinter(mem(z1, z0), z1, z2, z0)
   ifinter(true, z0, z1, z2) -> cons(z0, inter(z1, z2))
   ifinter(false, z0, z1, z2) -> inter(z1, z2)
Tuples:
   IF(true, z0, z1) -> c
   IF(false, z0, z1) -> c1
   EQ(0, 0) -> c2
   EQ(0, s(z0)) -> c3
   EQ(s(z0), 0) -> c4
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(nil, z0) -> c6
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, nil) -> c8
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(true, z0, z1) -> c10
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(z0, nil) -> c12
   INTER(nil, z0) -> c13
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
S tuples:
   IF(true, z0, z1) -> c
   IF(false, z0, z1) -> c1
   EQ(0, 0) -> c2
   EQ(0, s(z0)) -> c3
   EQ(s(z0), 0) -> c4
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(nil, z0) -> c6
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, nil) -> c8
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(true, z0, z1) -> c10
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(z0, nil) -> c12
   INTER(nil, z0) -> c13
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
K tuples:none
Defined Rule Symbols:   if_3, eq_2, app_2, mem_2, ifmem_3, inter_2, ifinter_4

Defined Pair Symbols:   IF_3, EQ_2, APP_2, MEM_2, IFMEM_3, INTER_2, IFINTER_4

Compound Symbols:   c, c1, c2, c3, c4, c5_1, c6, c7_1, c8, c9_2, c10, c11_1, c12, c13, c14_2, c15_2, c16_1, c17_1


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

(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 10 trailing nodes:
   INTER(nil, z0) -> c13
   APP(nil, z0) -> c6
   IF(true, z0, z1) -> c
   INTER(z0, nil) -> c12
   IFMEM(true, z0, z1) -> c10
   EQ(0, 0) -> c2
   MEM(z0, nil) -> c8
   EQ(0, s(z0)) -> c3
   EQ(s(z0), 0) -> c4
   IF(false, z0, z1) -> c1

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

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   if(true, z0, z1) -> z0
   if(false, z0, z1) -> z1
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   app(nil, z0) -> z0
   app(cons(z0, z1), z2) -> cons(z0, app(z1, z2))
   mem(z0, nil) -> false
   mem(z0, cons(z1, z2)) -> ifmem(eq(z0, z1), z0, z2)
   ifmem(true, z0, z1) -> true
   ifmem(false, z0, z1) -> mem(z0, z1)
   inter(z0, nil) -> nil
   inter(nil, z0) -> nil
   inter(cons(z0, z1), z2) -> ifinter(mem(z0, z2), z0, z1, z2)
   inter(z0, cons(z1, z2)) -> ifinter(mem(z1, z0), z1, z2, z0)
   ifinter(true, z0, z1, z2) -> cons(z0, inter(z1, z2))
   ifinter(false, z0, z1, z2) -> inter(z1, z2)
Tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
S tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
K tuples:none
Defined Rule Symbols:   if_3, eq_2, app_2, mem_2, ifmem_3, inter_2, ifinter_4

Defined Pair Symbols:   EQ_2, APP_2, MEM_2, IFMEM_3, INTER_2, IFINTER_4

Compound Symbols:   c5_1, c7_1, c9_2, c11_1, c14_2, c15_2, c16_1, c17_1


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

(9) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   if(true, z0, z1) -> z0
   if(false, z0, z1) -> z1
   app(nil, z0) -> z0
   app(cons(z0, z1), z2) -> cons(z0, app(z1, z2))
   inter(z0, nil) -> nil
   inter(nil, z0) -> nil
   inter(cons(z0, z1), z2) -> ifinter(mem(z0, z2), z0, z1, z2)
   inter(z0, cons(z1, z2)) -> ifinter(mem(z1, z0), z1, z2, z0)
   ifinter(true, z0, z1, z2) -> cons(z0, inter(z1, z2))
   ifinter(false, z0, z1, z2) -> inter(z1, z2)

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

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   mem(z0, nil) -> false
   mem(z0, cons(z1, z2)) -> ifmem(eq(z0, z1), z0, z2)
   ifmem(true, z0, z1) -> true
   ifmem(false, z0, z1) -> mem(z0, z1)
Tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
S tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
K tuples:none
Defined Rule Symbols:   eq_2, mem_2, ifmem_3

Defined Pair Symbols:   EQ_2, APP_2, MEM_2, IFMEM_3, INTER_2, IFINTER_4

Compound Symbols:   c5_1, c7_1, c9_2, c11_1, c14_2, c15_2, c16_1, c17_1


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

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
We considered the (Usable) Rules:
   mem(z0, cons(z1, z2)) -> ifmem(eq(z0, z1), z0, z2)
   ifmem(true, z0, z1) -> true
   ifmem(false, z0, z1) -> mem(z0, z1)
   mem(z0, nil) -> false
And the Tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [1]
   POL(APP(x_1, x_2)) = x_1
   POL(EQ(x_1, x_2)) = 0
   POL(IFINTER(x_1, x_2, x_3, x_4)) = [1] + x_1 + x_3 + x_4
   POL(IFMEM(x_1, x_2, x_3)) = 0
   POL(INTER(x_1, x_2)) = [1] + x_1 + x_2
   POL(MEM(x_1, x_2)) = 0
   POL(c11(x_1)) = x_1
   POL(c14(x_1, x_2)) = x_1 + x_2
   POL(c15(x_1, x_2)) = x_1 + x_2
   POL(c16(x_1)) = x_1
   POL(c17(x_1)) = x_1
   POL(c5(x_1)) = x_1
   POL(c7(x_1)) = x_1
   POL(c9(x_1, x_2)) = x_1 + x_2
   POL(cons(x_1, x_2)) = [1] + x_1 + x_2
   POL(eq(x_1, x_2)) = [1] + x_1 + x_2
   POL(false) = [1]
   POL(ifmem(x_1, x_2, x_3)) = [1] + x_2
   POL(mem(x_1, x_2)) = [1] + x_1
   POL(nil) = [1]
   POL(s(x_1)) = [1] + x_1
   POL(true) = [1]

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

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   mem(z0, nil) -> false
   mem(z0, cons(z1, z2)) -> ifmem(eq(z0, z1), z0, z2)
   ifmem(true, z0, z1) -> true
   ifmem(false, z0, z1) -> mem(z0, z1)
Tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
S tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
K tuples:
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
Defined Rule Symbols:   eq_2, mem_2, ifmem_3

Defined Pair Symbols:   EQ_2, APP_2, MEM_2, IFMEM_3, INTER_2, IFINTER_4

Compound Symbols:   c5_1, c7_1, c9_2, c11_1, c14_2, c15_2, c16_1, c17_1


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

(13) CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
The following tuples could be moved from S to K by knowledge propagation:
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))

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

(14)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   mem(z0, nil) -> false
   mem(z0, cons(z1, z2)) -> ifmem(eq(z0, z1), z0, z2)
   ifmem(true, z0, z1) -> true
   ifmem(false, z0, z1) -> mem(z0, z1)
Tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
S tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
K tuples:
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
Defined Rule Symbols:   eq_2, mem_2, ifmem_3

Defined Pair Symbols:   EQ_2, APP_2, MEM_2, IFMEM_3, INTER_2, IFINTER_4

Compound Symbols:   c5_1, c7_1, c9_2, c11_1, c14_2, c15_2, c16_1, c17_1


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

(15) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(APP(x_1, x_2)) = [2]x_1 + [2]x_1*x_2 + x_1^2
   POL(EQ(x_1, x_2)) = 0
   POL(IFINTER(x_1, x_2, x_3, x_4)) = x_4 + [2]x_3*x_4
   POL(IFMEM(x_1, x_2, x_3)) = x_3
   POL(INTER(x_1, x_2)) = [2]x_1*x_2
   POL(MEM(x_1, x_2)) = x_2
   POL(c11(x_1)) = x_1
   POL(c14(x_1, x_2)) = x_1 + x_2
   POL(c15(x_1, x_2)) = x_1 + x_2
   POL(c16(x_1)) = x_1
   POL(c17(x_1)) = x_1
   POL(c5(x_1)) = x_1
   POL(c7(x_1)) = x_1
   POL(c9(x_1, x_2)) = x_1 + x_2
   POL(cons(x_1, x_2)) = [2] + x_2
   POL(eq(x_1, x_2)) = [2]
   POL(false) = 0
   POL(ifmem(x_1, x_2, x_3)) = [2]x_1 + x_3 + [2]x_3^2
   POL(mem(x_1, x_2)) = x_2 + [2]x_2^2
   POL(nil) = 0
   POL(s(x_1)) = 0
   POL(true) = 0

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

(16)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   mem(z0, nil) -> false
   mem(z0, cons(z1, z2)) -> ifmem(eq(z0, z1), z0, z2)
   ifmem(true, z0, z1) -> true
   ifmem(false, z0, z1) -> mem(z0, z1)
Tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
S tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
K tuples:
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
Defined Rule Symbols:   eq_2, mem_2, ifmem_3

Defined Pair Symbols:   EQ_2, APP_2, MEM_2, IFMEM_3, INTER_2, IFINTER_4

Compound Symbols:   c5_1, c7_1, c9_2, c11_1, c14_2, c15_2, c16_1, c17_1


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

(17) CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
The following tuples could be moved from S to K by knowledge propagation:
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))

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

(18)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   mem(z0, nil) -> false
   mem(z0, cons(z1, z2)) -> ifmem(eq(z0, z1), z0, z2)
   ifmem(true, z0, z1) -> true
   ifmem(false, z0, z1) -> mem(z0, z1)
Tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
S tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
K tuples:
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
Defined Rule Symbols:   eq_2, mem_2, ifmem_3

Defined Pair Symbols:   EQ_2, APP_2, MEM_2, IFMEM_3, INTER_2, IFINTER_4

Compound Symbols:   c5_1, c7_1, c9_2, c11_1, c14_2, c15_2, c16_1, c17_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(s(z0), s(z1)) -> c5(EQ(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(APP(x_1, x_2)) = [2]x_1 + [2]x_1*x_2 + [2]x_1^2
   POL(EQ(x_1, x_2)) = x_1*x_2
   POL(IFINTER(x_1, x_2, x_3, x_4)) = [2]x_4 + [2]x_3*x_4
   POL(IFMEM(x_1, x_2, x_3)) = [2]x_2*x_3
   POL(INTER(x_1, x_2)) = [2]x_2 + [2]x_1*x_2
   POL(MEM(x_1, x_2)) = [2]x_1*x_2
   POL(c11(x_1)) = x_1
   POL(c14(x_1, x_2)) = x_1 + x_2
   POL(c15(x_1, x_2)) = x_1 + x_2
   POL(c16(x_1)) = x_1
   POL(c17(x_1)) = x_1
   POL(c5(x_1)) = x_1
   POL(c7(x_1)) = x_1
   POL(c9(x_1, x_2)) = x_1 + x_2
   POL(cons(x_1, x_2)) = [1] + x_1 + x_2
   POL(eq(x_1, x_2)) = [1]
   POL(false) = [2]
   POL(ifmem(x_1, x_2, x_3)) = [2]x_1 + [2]x_3^2
   POL(mem(x_1, x_2)) = [2] + [2]x_2^2
   POL(nil) = 0
   POL(s(x_1)) = [1] + x_1
   POL(true) = 0

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

(20)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   mem(z0, nil) -> false
   mem(z0, cons(z1, z2)) -> ifmem(eq(z0, z1), z0, z2)
   ifmem(true, z0, z1) -> true
   ifmem(false, z0, z1) -> mem(z0, z1)
Tuples:
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
S tuples:none
K tuples:
   APP(cons(z0, z1), z2) -> c7(APP(z1, z2))
   IFINTER(true, z0, z1, z2) -> c16(INTER(z1, z2))
   IFINTER(false, z0, z1, z2) -> c17(INTER(z1, z2))
   INTER(cons(z0, z1), z2) -> c14(IFINTER(mem(z0, z2), z0, z1, z2), MEM(z0, z2))
   INTER(z0, cons(z1, z2)) -> c15(IFINTER(mem(z1, z0), z1, z2, z0), MEM(z1, z0))
   MEM(z0, cons(z1, z2)) -> c9(IFMEM(eq(z0, z1), z0, z2), EQ(z0, z1))
   IFMEM(false, z0, z1) -> c11(MEM(z0, z1))
   EQ(s(z0), s(z1)) -> c5(EQ(z0, z1))
Defined Rule Symbols:   eq_2, mem_2, ifmem_3

Defined Pair Symbols:   EQ_2, APP_2, MEM_2, IFMEM_3, INTER_2, IFINTER_4

Compound Symbols:   c5_1, c7_1, c9_2, c11_1, c14_2, c15_2, c16_1, c17_1


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

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

(22)
BOUNDS(1, 1)

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

(23) RenamingProof (BOTH BOUNDS(ID, ID))
Renamed function symbols to avoid clashes with predefined symbol.
----------------------------------------

(24)
Obligation:
The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   if(true, x, y) -> x
   if(false, x, y) -> y
   eq(0', 0') -> true
   eq(0', s(x)) -> false
   eq(s(x), 0') -> false
   eq(s(x), s(y)) -> eq(x, y)
   app(nil, l) -> l
   app(cons(x, l1), l2) -> cons(x, app(l1, l2))
   app(app(l1, l2), l3) -> app(l1, app(l2, l3))
   mem(x, nil) -> false
   mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
   ifmem(true, x, l) -> true
   ifmem(false, x, l) -> 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)

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

(25) TypeInferenceProof (BOTH BOUNDS(ID, ID))
Infered types.
----------------------------------------

(26)
Obligation:
TRS:
Rules:
if(true, x, y) -> x
if(false, x, y) -> y
eq(0', 0') -> true
eq(0', s(x)) -> false
eq(s(x), 0') -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> 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)

Types:
if :: true:false -> if -> if -> if
true :: true:false
false :: true:false
eq :: 0':s -> 0':s -> true:false
0' :: 0':s
s :: 0':s -> 0':s
app :: nil:cons -> nil:cons -> nil:cons
nil :: nil:cons
cons :: 0':s -> nil:cons -> nil:cons
mem :: 0':s -> nil:cons -> true:false
ifmem :: true:false -> 0':s -> nil:cons -> true:false
inter :: nil:cons -> nil:cons -> nil:cons
ifinter :: true:false -> 0':s -> nil:cons -> nil:cons -> nil:cons
hole_if1_0 :: if
hole_true:false2_0 :: true:false
hole_0':s3_0 :: 0':s
hole_nil:cons4_0 :: nil:cons
gen_0':s5_0 :: Nat -> 0':s
gen_nil:cons6_0 :: Nat -> nil:cons

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

(27) OrderProof (LOWER BOUND(ID))
Heuristically decided to analyse the following defined symbols:
eq, app, mem, inter

They will be analysed ascendingly in the following order:
eq < mem
app < inter
mem < inter

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

(28)
Obligation:
TRS:
Rules:
if(true, x, y) -> x
if(false, x, y) -> y
eq(0', 0') -> true
eq(0', s(x)) -> false
eq(s(x), 0') -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> 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)

Types:
if :: true:false -> if -> if -> if
true :: true:false
false :: true:false
eq :: 0':s -> 0':s -> true:false
0' :: 0':s
s :: 0':s -> 0':s
app :: nil:cons -> nil:cons -> nil:cons
nil :: nil:cons
cons :: 0':s -> nil:cons -> nil:cons
mem :: 0':s -> nil:cons -> true:false
ifmem :: true:false -> 0':s -> nil:cons -> true:false
inter :: nil:cons -> nil:cons -> nil:cons
ifinter :: true:false -> 0':s -> nil:cons -> nil:cons -> nil:cons
hole_if1_0 :: if
hole_true:false2_0 :: true:false
hole_0':s3_0 :: 0':s
hole_nil:cons4_0 :: nil:cons
gen_0':s5_0 :: Nat -> 0':s
gen_nil:cons6_0 :: Nat -> nil:cons


Generator Equations:
gen_0':s5_0(0) <=> 0'
gen_0':s5_0(+(x, 1)) <=> s(gen_0':s5_0(x))
gen_nil:cons6_0(0) <=> nil
gen_nil:cons6_0(+(x, 1)) <=> cons(0', gen_nil:cons6_0(x))


The following defined symbols remain to be analysed:
eq, app, mem, inter

They will be analysed ascendingly in the following order:
eq < mem
app < inter
mem < inter

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

(29) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
eq(gen_0':s5_0(n8_0), gen_0':s5_0(n8_0)) -> true, rt in Omega(1 + n8_0)

Induction Base:
eq(gen_0':s5_0(0), gen_0':s5_0(0)) ->_R^Omega(1)
true

Induction Step:
eq(gen_0':s5_0(+(n8_0, 1)), gen_0':s5_0(+(n8_0, 1))) ->_R^Omega(1)
eq(gen_0':s5_0(n8_0), gen_0':s5_0(n8_0)) ->_IH
true

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(30)
Complex Obligation (BEST)

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

(31)
Obligation:
Proved the lower bound n^1 for the following obligation:

TRS:
Rules:
if(true, x, y) -> x
if(false, x, y) -> y
eq(0', 0') -> true
eq(0', s(x)) -> false
eq(s(x), 0') -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> 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)

Types:
if :: true:false -> if -> if -> if
true :: true:false
false :: true:false
eq :: 0':s -> 0':s -> true:false
0' :: 0':s
s :: 0':s -> 0':s
app :: nil:cons -> nil:cons -> nil:cons
nil :: nil:cons
cons :: 0':s -> nil:cons -> nil:cons
mem :: 0':s -> nil:cons -> true:false
ifmem :: true:false -> 0':s -> nil:cons -> true:false
inter :: nil:cons -> nil:cons -> nil:cons
ifinter :: true:false -> 0':s -> nil:cons -> nil:cons -> nil:cons
hole_if1_0 :: if
hole_true:false2_0 :: true:false
hole_0':s3_0 :: 0':s
hole_nil:cons4_0 :: nil:cons
gen_0':s5_0 :: Nat -> 0':s
gen_nil:cons6_0 :: Nat -> nil:cons


Generator Equations:
gen_0':s5_0(0) <=> 0'
gen_0':s5_0(+(x, 1)) <=> s(gen_0':s5_0(x))
gen_nil:cons6_0(0) <=> nil
gen_nil:cons6_0(+(x, 1)) <=> cons(0', gen_nil:cons6_0(x))


The following defined symbols remain to be analysed:
eq, app, mem, inter

They will be analysed ascendingly in the following order:
eq < mem
app < inter
mem < inter

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

(32) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
----------------------------------------

(33)
BOUNDS(n^1, INF)

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

(34)
Obligation:
TRS:
Rules:
if(true, x, y) -> x
if(false, x, y) -> y
eq(0', 0') -> true
eq(0', s(x)) -> false
eq(s(x), 0') -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> 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)

Types:
if :: true:false -> if -> if -> if
true :: true:false
false :: true:false
eq :: 0':s -> 0':s -> true:false
0' :: 0':s
s :: 0':s -> 0':s
app :: nil:cons -> nil:cons -> nil:cons
nil :: nil:cons
cons :: 0':s -> nil:cons -> nil:cons
mem :: 0':s -> nil:cons -> true:false
ifmem :: true:false -> 0':s -> nil:cons -> true:false
inter :: nil:cons -> nil:cons -> nil:cons
ifinter :: true:false -> 0':s -> nil:cons -> nil:cons -> nil:cons
hole_if1_0 :: if
hole_true:false2_0 :: true:false
hole_0':s3_0 :: 0':s
hole_nil:cons4_0 :: nil:cons
gen_0':s5_0 :: Nat -> 0':s
gen_nil:cons6_0 :: Nat -> nil:cons


Lemmas:
eq(gen_0':s5_0(n8_0), gen_0':s5_0(n8_0)) -> true, rt in Omega(1 + n8_0)


Generator Equations:
gen_0':s5_0(0) <=> 0'
gen_0':s5_0(+(x, 1)) <=> s(gen_0':s5_0(x))
gen_nil:cons6_0(0) <=> nil
gen_nil:cons6_0(+(x, 1)) <=> cons(0', gen_nil:cons6_0(x))


The following defined symbols remain to be analysed:
app, mem, inter

They will be analysed ascendingly in the following order:
app < inter
mem < inter

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

(35) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
app(gen_nil:cons6_0(n537_0), gen_nil:cons6_0(b)) -> gen_nil:cons6_0(+(n537_0, b)), rt in Omega(1 + n537_0)

Induction Base:
app(gen_nil:cons6_0(0), gen_nil:cons6_0(b)) ->_R^Omega(1)
gen_nil:cons6_0(b)

Induction Step:
app(gen_nil:cons6_0(+(n537_0, 1)), gen_nil:cons6_0(b)) ->_R^Omega(1)
cons(0', app(gen_nil:cons6_0(n537_0), gen_nil:cons6_0(b))) ->_IH
cons(0', gen_nil:cons6_0(+(b, c538_0)))

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(36)
Obligation:
TRS:
Rules:
if(true, x, y) -> x
if(false, x, y) -> y
eq(0', 0') -> true
eq(0', s(x)) -> false
eq(s(x), 0') -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> 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)

Types:
if :: true:false -> if -> if -> if
true :: true:false
false :: true:false
eq :: 0':s -> 0':s -> true:false
0' :: 0':s
s :: 0':s -> 0':s
app :: nil:cons -> nil:cons -> nil:cons
nil :: nil:cons
cons :: 0':s -> nil:cons -> nil:cons
mem :: 0':s -> nil:cons -> true:false
ifmem :: true:false -> 0':s -> nil:cons -> true:false
inter :: nil:cons -> nil:cons -> nil:cons
ifinter :: true:false -> 0':s -> nil:cons -> nil:cons -> nil:cons
hole_if1_0 :: if
hole_true:false2_0 :: true:false
hole_0':s3_0 :: 0':s
hole_nil:cons4_0 :: nil:cons
gen_0':s5_0 :: Nat -> 0':s
gen_nil:cons6_0 :: Nat -> nil:cons


Lemmas:
eq(gen_0':s5_0(n8_0), gen_0':s5_0(n8_0)) -> true, rt in Omega(1 + n8_0)
app(gen_nil:cons6_0(n537_0), gen_nil:cons6_0(b)) -> gen_nil:cons6_0(+(n537_0, b)), rt in Omega(1 + n537_0)


Generator Equations:
gen_0':s5_0(0) <=> 0'
gen_0':s5_0(+(x, 1)) <=> s(gen_0':s5_0(x))
gen_nil:cons6_0(0) <=> nil
gen_nil:cons6_0(+(x, 1)) <=> cons(0', gen_nil:cons6_0(x))


The following defined symbols remain to be analysed:
mem, inter

They will be analysed ascendingly in the following order:
mem < inter