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

0 CpxTRS
↳1 NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID), 3 ms)
↳2 CpxTRS
    ↳3 CpxTrsMatchBoundsTAProof (⇔, 66 ms)
    ↳4 BOUNDS(1, n^1)
    ↳5 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳6 CdtProblem
        ↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳10 CdtProblem
        ↳11 CdtUsableRulesProof (⇔, 0 ms)
        ↳12 CdtProblem
        ↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 103 ms)
        ↳14 CdtProblem
        ↳15 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 10 ms)
        ↳16 CdtProblem
        ↳17 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 18 ms)
        ↳18 CdtProblem
        ↳19 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 23 ms)
        ↳20 CdtProblem
        ↳21 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 28 ms)
        ↳22 CdtProblem
        ↳23 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳24 BOUNDS(1, 1)

(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(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
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(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))

(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:

div(mark(X1), X2) → mark(div(X1, X2))
proper(true) → ok(true)
top(ok(X)) → top(active(X))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
s(ok(X)) → ok(s(X))
s(mark(X)) → mark(s(X))
proper(false) → ok(false)
proper(0) → ok(0)
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))

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 2.

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, 6, 7]
transitions:
mark0(0) → 0
true0() → 0
ok0(0) → 0
active0(0) → 0
false0() → 0
00() → 0
div0(0, 0) → 1
proper0(0) → 2
top0(0) → 3
minus0(0, 0) → 4
s0(0) → 5
if0(0, 0, 0) → 6
geq0(0, 0) → 7
div1(0, 0) → 8
mark1(8) → 1
true1() → 9
ok1(9) → 2
active1(0) → 10
top1(10) → 3
minus1(0, 0) → 11
ok1(11) → 4
div1(0, 0) → 12
ok1(12) → 1
s1(0) → 13
ok1(13) → 5
s1(0) → 14
mark1(14) → 5
false1() → 15
ok1(15) → 2
01() → 16
ok1(16) → 2
if1(0, 0, 0) → 17
mark1(17) → 6
if1(0, 0, 0) → 18
ok1(18) → 6
proper1(0) → 19
top1(19) → 3
geq1(0, 0) → 20
ok1(20) → 7
mark1(8) → 8
mark1(8) → 12
ok1(9) → 19
ok1(11) → 11
ok1(12) → 8
ok1(12) → 12
ok1(13) → 13
ok1(13) → 14
mark1(14) → 13
mark1(14) → 14
ok1(15) → 19
ok1(16) → 19
mark1(17) → 17
mark1(17) → 18
ok1(18) → 17
ok1(18) → 18
ok1(20) → 20
active2(9) → 21
top2(21) → 3
active2(15) → 21
active2(16) → 21

(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:

div(mark(z0), z1) → mark(div(z0, z1))
div(ok(z0), ok(z1)) → ok(div(z0, z1))
proper(true) → ok(true)
proper(false) → ok(false)
proper(0) → ok(0)
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
minus(ok(z0), ok(z1)) → ok(minus(z0, z1))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
if(mark(z0), z1, z2) → mark(if(z0, z1, z2))
if(ok(z0), ok(z1), ok(z2)) → ok(if(z0, z1, z2))
geq(ok(z0), ok(z1)) → ok(geq(z0, z1))
Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
PROPER(true) → c2
PROPER(false) → c3
PROPER(0) → c4
TOP(ok(z0)) → c5(TOP(active(z0)))
TOP(mark(z0)) → c6(TOP(proper(z0)), PROPER(z0))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
S tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
PROPER(true) → c2
PROPER(false) → c3
PROPER(0) → c4
TOP(ok(z0)) → c5(TOP(active(z0)))
TOP(mark(z0)) → c6(TOP(proper(z0)), PROPER(z0))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
K tuples:none
Defined Rule Symbols:

div, proper, top, minus, s, if, geq

Defined Pair Symbols:

DIV, PROPER, TOP, MINUS, S, IF, GEQ

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12

(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 4 trailing nodes:

PROPER(false) → c3
PROPER(true) → c2
TOP(ok(z0)) → c5(TOP(active(z0)))
PROPER(0) → c4

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

div(mark(z0), z1) → mark(div(z0, z1))
div(ok(z0), ok(z1)) → ok(div(z0, z1))
proper(true) → ok(true)
proper(false) → ok(false)
proper(0) → ok(0)
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
minus(ok(z0), ok(z1)) → ok(minus(z0, z1))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
if(mark(z0), z1, z2) → mark(if(z0, z1, z2))
if(ok(z0), ok(z1), ok(z2)) → ok(if(z0, z1, z2))
geq(ok(z0), ok(z1)) → ok(geq(z0, z1))
Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)), PROPER(z0))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
S tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)), PROPER(z0))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
K tuples:none
Defined Rule Symbols:

div, proper, top, minus, s, if, geq

Defined Pair Symbols:

DIV, TOP, MINUS, S, IF, GEQ

Compound Symbols:

c, c1, c6, c7, c8, c9, c10, c11, c12

(9) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

div(mark(z0), z1) → mark(div(z0, z1))
div(ok(z0), ok(z1)) → ok(div(z0, z1))
proper(true) → ok(true)
proper(false) → ok(false)
proper(0) → ok(0)
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
minus(ok(z0), ok(z1)) → ok(minus(z0, z1))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
if(mark(z0), z1, z2) → mark(if(z0, z1, z2))
if(ok(z0), ok(z1), ok(z2)) → ok(if(z0, z1, z2))
geq(ok(z0), ok(z1)) → ok(geq(z0, z1))
Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

div, proper, top, minus, s, if, geq

Defined Pair Symbols:

DIV, MINUS, S, IF, GEQ, TOP

Compound Symbols:

c, c1, c7, c8, c9, c10, c11, c12, c6

(11) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

div(mark(z0), z1) → mark(div(z0, z1))
div(ok(z0), ok(z1)) → ok(div(z0, z1))
top(ok(z0)) → top(active(z0))
top(mark(z0)) → top(proper(z0))
minus(ok(z0), ok(z1)) → ok(minus(z0, z1))
s(ok(z0)) → ok(s(z0))
s(mark(z0)) → mark(s(z0))
if(mark(z0), z1, z2) → mark(if(z0, z1, z2))
if(ok(z0), ok(z1), ok(z2)) → ok(if(z0, z1, z2))
geq(ok(z0), ok(z1)) → ok(geq(z0, z1))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
K tuples:none
Defined Rule Symbols:

proper

Defined Pair Symbols:

DIV, MINUS, S, IF, GEQ, TOP

Compound Symbols:

c, c1, c7, c8, c9, c10, c11, c12, c6

(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.

DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(DIV(x1, x2)) = x2   
POL(GEQ(x1, x2)) = x2   
POL(IF(x1, x2, x3)) = 0   
POL(MINUS(x1, x2)) = x1   
POL(S(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = 0   
POL(ok(x1)) = [1] + x1   
POL(proper(x1)) = 0   
POL(true) = 0   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
TOP(mark(z0)) → c6(TOP(proper(z0)))
K tuples:

DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
Defined Rule Symbols:

proper

Defined Pair Symbols:

DIV, MINUS, S, IF, GEQ, TOP

Compound Symbols:

c, c1, c7, c8, c9, c10, c11, c12, c6

(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.

DIV(mark(z0), z1) → c(DIV(z0, z1))
S(mark(z0)) → c9(S(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
We considered the (Usable) Rules:

proper(true) → ok(true)
proper(false) → ok(false)
proper(0) → ok(0)
And the Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(DIV(x1, x2)) = x1   
POL(GEQ(x1, x2)) = 0   
POL(IF(x1, x2, x3)) = 0   
POL(MINUS(x1, x2)) = 0   
POL(S(x1)) = x1   
POL(TOP(x1)) = x1   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = x1   
POL(proper(x1)) = 0   
POL(true) = 0   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

S(ok(z0)) → c8(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
K tuples:

DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
DIV(mark(z0), z1) → c(DIV(z0, z1))
S(mark(z0)) → c9(S(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
Defined Rule Symbols:

proper

Defined Pair Symbols:

DIV, MINUS, S, IF, GEQ, TOP

Compound Symbols:

c, c1, c7, c8, c9, c10, c11, c12, c6

(17) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(DIV(x1, x2)) = 0   
POL(GEQ(x1, x2)) = 0   
POL(IF(x1, x2, x3)) = x1   
POL(MINUS(x1, x2)) = 0   
POL(S(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = [1] + x1   
POL(ok(x1)) = x1   
POL(proper(x1)) = 0   
POL(true) = 0   

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

S(ok(z0)) → c8(S(z0))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
K tuples:

DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
DIV(mark(z0), z1) → c(DIV(z0, z1))
S(mark(z0)) → c9(S(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
Defined Rule Symbols:

proper

Defined Pair Symbols:

DIV, MINUS, S, IF, GEQ, TOP

Compound Symbols:

c, c1, c7, c8, c9, c10, c11, c12, c6

(19) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(DIV(x1, x2)) = x2   
POL(GEQ(x1, x2)) = 0   
POL(IF(x1, x2, x3)) = [2]x3   
POL(MINUS(x1, x2)) = x1   
POL(S(x1)) = 0   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = 0   
POL(ok(x1)) = [2] + x1   
POL(proper(x1)) = 0   
POL(true) = 0   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:

S(ok(z0)) → c8(S(z0))
K tuples:

DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
DIV(mark(z0), z1) → c(DIV(z0, z1))
S(mark(z0)) → c9(S(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
Defined Rule Symbols:

proper

Defined Pair Symbols:

DIV, MINUS, S, IF, GEQ, TOP

Compound Symbols:

c, c1, c7, c8, c9, c10, c11, c12, c6

(21) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

S(ok(z0)) → c8(S(z0))
We considered the (Usable) Rules:none
And the Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(DIV(x1, x2)) = 0   
POL(GEQ(x1, x2)) = x1   
POL(IF(x1, x2, x3)) = [2]x2 + [2]x3   
POL(MINUS(x1, x2)) = 0   
POL(S(x1)) = [2]x1   
POL(TOP(x1)) = 0   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c10(x1)) = x1   
POL(c11(x1)) = x1   
POL(c12(x1)) = x1   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(c8(x1)) = x1   
POL(c9(x1)) = x1   
POL(false) = 0   
POL(mark(x1)) = x1   
POL(ok(x1)) = [2] + x1   
POL(proper(x1)) = 0   
POL(true) = 0   

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

proper(true) → ok(true)
proper(false) → ok(false)
proper(0) → ok(0)
Tuples:

DIV(mark(z0), z1) → c(DIV(z0, z1))
DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
S(ok(z0)) → c8(S(z0))
S(mark(z0)) → c9(S(z0))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
TOP(mark(z0)) → c6(TOP(proper(z0)))
S tuples:none
K tuples:

DIV(ok(z0), ok(z1)) → c1(DIV(z0, z1))
MINUS(ok(z0), ok(z1)) → c7(MINUS(z0, z1))
GEQ(ok(z0), ok(z1)) → c12(GEQ(z0, z1))
DIV(mark(z0), z1) → c(DIV(z0, z1))
S(mark(z0)) → c9(S(z0))
TOP(mark(z0)) → c6(TOP(proper(z0)))
IF(mark(z0), z1, z2) → c10(IF(z0, z1, z2))
IF(ok(z0), ok(z1), ok(z2)) → c11(IF(z0, z1, z2))
S(ok(z0)) → c8(S(z0))
Defined Rule Symbols:

proper

Defined Pair Symbols:

DIV, MINUS, S, IF, GEQ, TOP

Compound Symbols:

c, c1, c7, c8, c9, c10, c11, c12, c6

(23) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(24) BOUNDS(1, 1)