Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2, X3]
active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACTIVE(f(X)) -> IF(X, c, f(true))
ACTIVE(f(X)) -> F(true)
ACTIVE(f(X)) -> F(active(X))
ACTIVE(f(X)) -> ACTIVE(X)
ACTIVE(if(X1, X2, X3)) -> IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) -> IF(X1, active(X2), X3)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X2)
F(mark(X)) -> F(X)
F(ok(X)) -> F(X)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)
IF(X1, mark(X2), X3) -> IF(X1, X2, X3)
IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
PROPER(f(X)) -> F(proper(X))
PROPER(f(X)) -> PROPER(X)
PROPER(if(X1, X2, X3)) -> IF(proper(X1), proper(X2), proper(X3))
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X3)
TOP(mark(X)) -> TOP(proper(X))
TOP(mark(X)) -> PROPER(X)
TOP(ok(X)) -> TOP(active(X))
TOP(ok(X)) -> ACTIVE(X)

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

F(ok(X)) -> F(X)
F(mark(X)) -> F(X)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 18 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 6

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

F(ok(X)) -> F(X)
F(mark(X)) -> F(X)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

F(ok(X)) -> F(X)
F(mark(X)) -> F(X)

and get the following Size-Change Graph(s):

{1, 2}	,	{1, 2}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1, 2}	,	{1, 2}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
IF(X1, mark(X2), X3) -> IF(X1, X2, X3)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 18 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 7

             ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
IF(X1, mark(X2), X3) -> IF(X1, X2, X3)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
IF(X1, mark(X2), X3) -> IF(X1, X2, X3)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)

and get the following Size-Change Graph(s):

{1, 2, 3}	,	{1, 2, 3}
1	>	1
2	>	2
3	>	3

{1, 2, 3}	,	{1, 2, 3}
1	=	1
2	>	2
3	=	3

{1, 2, 3}	,	{1, 2, 3}
1	>	1
2	=	2
3	=	3

which lead(s) to this/these maximal multigraph(s):

{1, 2, 3}	,	{1, 2, 3}
1	>	1
2	=	2
3	=	3

{1, 2, 3}	,	{1, 2, 3}
1	=	1
2	>	2
3	=	3

{1, 2, 3}	,	{1, 2, 3}
1	>	1
2	>	2
3	>	3

{1, 2, 3}	,	{1, 2, 3}
1	>	1
2	>	2
3	=	3

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

ACTIVE(if(X1, X2, X3)) -> ACTIVE(X2)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(f(X)) -> ACTIVE(X)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 18 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 8

             ↳Size-Change Principle

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

ACTIVE(if(X1, X2, X3)) -> ACTIVE(X2)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(f(X)) -> ACTIVE(X)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

ACTIVE(if(X1, X2, X3)) -> ACTIVE(X2)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(f(X)) -> ACTIVE(X)

and get the following Size-Change Graph(s):

{1, 2, 3}	,	{1, 2, 3}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1, 2, 3}	,	{1, 2, 3}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

if(x₁, x₂, x₃) -> if(x₁, x₂, x₃)
f(x₁) -> f(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳Usable Rules (Innermost)

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 18 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

           →DP Problem 9

             ↳Size-Change Principle

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(f(X)) -> PROPER(X)

and get the following Size-Change Graph(s):

{1, 2, 3, 4}	,	{1, 2, 3, 4}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1, 2, 3, 4}	,	{1, 2, 3, 4}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

if(x₁, x₂, x₃) -> if(x₁, x₂, x₃)
f(x₁) -> f(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳Usable Rules (Innermost)

Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(f(X)) -> mark(if(X, c, f(true)))
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
f(mark(X)) -> mark(f(X))
f(ok(X)) -> ok(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
proper(f(X)) -> f(proper(X))
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(c) -> ok(c)
proper(true) -> ok(true)
proper(false) -> ok(false)
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳Modular Removal of Rules

Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> mark(if(X, c, f(true)))
f(ok(X)) -> ok(f(X))
f(mark(X)) -> mark(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(true) -> ok(true)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(false) -> ok(false)

Strategy:

innermost

We have the following set of usable rules:

active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
active(if(true, X, Y)) -> mark(X)
active(if(false, X, Y)) -> mark(Y)
active(f(X)) -> mark(if(X, c, f(true)))
f(ok(X)) -> ok(f(X))
f(mark(X)) -> mark(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(true) -> ok(true)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(false) -> ok(false)

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(active(x₁)) = x₁

POL(proper(x₁)) = x₁

POL(if(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(c) = 0

POL(false) = 1

POL(true) = 0

POL(mark(x₁)) = x₁

POL(TOP(x₁)) = x₁

POL(ok(x₁)) = x₁

POL(f(x₁)) = x₁

We have the following set D of usable symbols: {proper, active, c, if, false, true, mark, TOP, ok, f}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

active(if(false, X, Y)) -> mark(Y)

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 11

                 ↳Negative Polynomial Order

Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
active(if(true, X, Y)) -> mark(X)
active(f(X)) -> mark(if(X, c, f(true)))
f(ok(X)) -> ok(f(X))
f(mark(X)) -> mark(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(true) -> ok(true)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(false) -> ok(false)

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

TOP(mark(X)) -> TOP(proper(X))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
f(ok(X)) -> ok(f(X))
f(mark(X)) -> mark(f(X))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
active(if(true, X, Y)) -> mark(X)
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
active(f(X)) -> mark(if(X, c, f(true)))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(true) -> ok(true)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(false) -> ok(false)

Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x₁) ) = x₁

POL( mark(x₁) ) = x₁ + 1

POL( proper(x₁) ) = x₁

POL( ok(x₁) ) = x₁

POL( active(x₁) ) = x₁

POL( f(x₁) ) = x₁ + 1

POL( if(x₁, ..., x₃) ) = x₁ + x₂

POL( true ) = 1

POL( c ) = 0

POL( false ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 12

                 ↳Usable Rules (Innermost)

Dependency Pair:

TOP(ok(X)) -> TOP(active(X))

Rules:

active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
active(if(true, X, Y)) -> mark(X)
active(f(X)) -> mark(if(X, c, f(true)))
f(ok(X)) -> ok(f(X))
f(mark(X)) -> mark(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(true) -> ok(true)
proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
proper(false) -> ok(false)

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 13

                 ↳Modular Removal of Rules

Dependency Pair:

TOP(ok(X)) -> TOP(active(X))

Rules:

active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
active(if(true, X, Y)) -> mark(X)
active(f(X)) -> mark(if(X, c, f(true)))
f(ok(X)) -> ok(f(X))
f(mark(X)) -> mark(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))

Strategy:

innermost

We have the following set of usable rules:

active(f(X)) -> f(active(X))
active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
active(if(true, X, Y)) -> mark(X)
active(f(X)) -> mark(if(X, c, f(true)))
f(ok(X)) -> ok(f(X))
f(mark(X)) -> mark(f(X))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(active(x₁)) = x₁

POL(if(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(c) = 0

POL(true) = 0

POL(mark(x₁)) = x₁

POL(TOP(x₁)) = 1 + x₁

POL(ok(x₁)) = 1 + x₁

POL(f(x₁)) = x₁

We have the following set D of usable symbols: {active, c, if, true, mark, TOP, ok, f}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

TOP(ok(X)) -> TOP(active(X))

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

Innermost Termination of R successfully shown.
Duration:
0:04 minutes

POL(active(x₁))	= x₁
POL(proper(x₁))	= x₁
POL(if(x₁, x₂, x₃))	= x₁ + x₂ + x₃
POL(c)	= 0
POL(false)	= 1
POL(true)	= 0
POL(mark(x₁))	= x₁
POL(TOP(x₁))	= x₁
POL(ok(x₁))	= x₁
POL(f(x₁))	= x₁