(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^1).
The TRS R consists of the following rules:
active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Rewrite Strategy: INNERMOST
(1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)
The following defined symbols can occur below the 0th argument of top: proper, active
The following defined symbols can occur below the 0th argument of proper: proper, active
The following defined symbols can occur below the 0th argument of active: proper, active
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
(2) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^1).
The TRS R consists of the following rules:
top(ok(X)) → top(active(X))
from(ok(X)) → ok(from(X))
from(mark(X)) → mark(from(X))
s(ok(X)) → ok(s(X))
s(mark(X)) → mark(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
2nd(ok(X)) → ok(2nd(X))
top(mark(X)) → top(proper(X))
Rewrite Strategy: INNERMOST
(3) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)
A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 1.
The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3, 4, 5]
transitions:
ok0(0) → 0
active0(0) → 0
mark0(0) → 0
proper0(0) → 0
top0(0) → 1
from0(0) → 2
s0(0) → 3
cons0(0, 0) → 4
2nd0(0) → 5
active1(0) → 6
top1(6) → 1
from1(0) → 7
ok1(7) → 2
from1(0) → 8
mark1(8) → 2
s1(0) → 9
ok1(9) → 3
s1(0) → 10
mark1(10) → 3
cons1(0, 0) → 11
ok1(11) → 4
2nd1(0) → 12
mark1(12) → 5
cons1(0, 0) → 13
mark1(13) → 4
2nd1(0) → 14
ok1(14) → 5
proper1(0) → 15
top1(15) → 1
ok1(7) → 7
ok1(7) → 8
mark1(8) → 7
mark1(8) → 8
ok1(9) → 9
ok1(9) → 10
mark1(10) → 9
mark1(10) → 10
ok1(11) → 11
ok1(11) → 13
mark1(12) → 12
mark1(12) → 14
mark1(13) → 11
mark1(13) → 13
ok1(14) → 12
ok1(14) → 14
(4) BOUNDS(1, n^1)
(5) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
from(ok(z0)) → ok(from(z0))
from(mark(z0)) → mark(from(z0))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(z0, z1))
2nd(mark(z0)) → mark(2nd(z0))
2nd(ok(z0)) → ok(2nd(z0))
Tuples:
TOP(ok(z0)) → c(TOP(active(z0)))
TOP(mark(z0)) → c1(TOP(proper(z0)))
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
S tuples:
TOP(ok(z0)) → c(TOP(active(z0)))
TOP(mark(z0)) → c1(TOP(proper(z0)))
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
K tuples:none
Defined Rule Symbols:
top, from, s, cons, 2nd
Defined Pair Symbols:
TOP, FROM, S, CONS, 2ND
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7, c8, c9
(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
TOP(mark(z0)) → c1(TOP(proper(z0)))
TOP(ok(z0)) → c(TOP(active(z0)))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
from(ok(z0)) → ok(from(z0))
from(mark(z0)) → mark(from(z0))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(z0, z1))
2nd(mark(z0)) → mark(2nd(z0))
2nd(ok(z0)) → ok(2nd(z0))
Tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
S tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
K tuples:none
Defined Rule Symbols:
top, from, s, cons, 2nd
Defined Pair Symbols:
FROM, S, CONS, 2ND
Compound Symbols:
c2, c3, c4, c5, c6, c7, c8, c9
(9) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
from(ok(z0)) → ok(from(z0))
from(mark(z0)) → mark(from(z0))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
cons(ok(z0), ok(z1)) → ok(cons(z0, z1))
cons(mark(z0), z1) → mark(cons(z0, z1))
2nd(mark(z0)) → mark(2nd(z0))
2nd(ok(z0)) → ok(2nd(z0))
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
S tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
FROM, S, CONS, 2ND
Compound Symbols:
c2, c3, c4, c5, c6, c7, c8, c9
(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(2ND(x1)) = 0
POL(CONS(x1, x2)) = [2]x2
POL(FROM(x1)) = [2]x12
POL(S(x1)) = [2]x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(c9(x1)) = x1
POL(mark(x1)) = [2] + x1
POL(ok(x1)) = [2] + x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
S tuples:
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
K tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
Defined Rule Symbols:none
Defined Pair Symbols:
FROM, S, CONS, 2ND
Compound Symbols:
c2, c3, c4, c5, c6, c7, c8, c9
(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
2ND(ok(z0)) → c9(2ND(z0))
We considered the (Usable) Rules:none
And the Tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(2ND(x1)) = [2]x1
POL(CONS(x1, x2)) = 0
POL(FROM(x1)) = 0
POL(S(x1)) = 0
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(c9(x1)) = x1
POL(mark(x1)) = x1
POL(ok(x1)) = [2] + x1
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
S tuples:
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
K tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
2ND(ok(z0)) → c9(2ND(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
FROM, S, CONS, 2ND
Compound Symbols:
c2, c3, c4, c5, c6, c7, c8, c9
(15) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
We considered the (Usable) Rules:none
And the Tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(2ND(x1)) = x1
POL(CONS(x1, x2)) = x1
POL(FROM(x1)) = 0
POL(S(x1)) = 0
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c4(x1)) = x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(c9(x1)) = x1
POL(mark(x1)) = [1] + x1
POL(ok(x1)) = x1
(16) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
2ND(ok(z0)) → c9(2ND(z0))
S tuples:none
K tuples:
FROM(ok(z0)) → c2(FROM(z0))
FROM(mark(z0)) → c3(FROM(z0))
S(ok(z0)) → c4(S(z0))
S(mark(z0)) → c5(S(z0))
CONS(ok(z0), ok(z1)) → c6(CONS(z0, z1))
2ND(ok(z0)) → c9(2ND(z0))
CONS(mark(z0), z1) → c7(CONS(z0, z1))
2ND(mark(z0)) → c8(2ND(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
FROM, S, CONS, 2ND
Compound Symbols:
c2, c3, c4, c5, c6, c7, c8, c9
(17) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(18) BOUNDS(1, 1)