Termination Proof using AProVE

Term Rewriting System R:
[Y, X, X1, X2, X3]
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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
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(minus(s(X), s(Y))) -> MINUS(X, Y)
ACTIVE(geq(s(X), s(Y))) -> GEQ(X, Y)
ACTIVE(div(s(X), s(Y))) -> IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
ACTIVE(div(s(X), s(Y))) -> GEQ(X, Y)
ACTIVE(div(s(X), s(Y))) -> S(div(minus(X, Y), s(Y)))
ACTIVE(div(s(X), s(Y))) -> DIV(minus(X, Y), s(Y))
ACTIVE(div(s(X), s(Y))) -> MINUS(X, Y)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(div(X1, X2)) -> DIV(active(X1), X2)
ACTIVE(div(X1, X2)) -> ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) -> IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
DIV(mark(X1), X2) -> DIV(X1, X2)
DIV(ok(X1), ok(X2)) -> DIV(X1, X2)
IF(mark(X1), X2, X3) -> IF(X1, X2, X3)
IF(ok(X1), ok(X2), ok(X3)) -> IF(X1, X2, X3)
PROPER(minus(X1, X2)) -> MINUS(proper(X1), proper(X2))
PROPER(minus(X1, X2)) -> PROPER(X1)
PROPER(minus(X1, X2)) -> PROPER(X2)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
PROPER(geq(X1, X2)) -> GEQ(proper(X1), proper(X2))
PROPER(geq(X1, X2)) -> PROPER(X1)
PROPER(geq(X1, X2)) -> PROPER(X2)
PROPER(div(X1, X2)) -> DIV(proper(X1), proper(X2))
PROPER(div(X1, X2)) -> PROPER(X1)
PROPER(div(X1, X2)) -> PROPER(X2)
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)
MINUS(ok(X1), ok(X2)) -> MINUS(X1, X2)
GEQ(ok(X1), ok(X2)) -> GEQ(X1, X2)
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 eight SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pairs:

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

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

IF(x₁, x₂, x₃) -> IF(x₁, x₂, x₃)
ok(x₁) -> ok(x₁)
mark(x₁) -> mark(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pair:

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pairs:

S(ok(X)) -> S(X)
S(mark(X)) -> S(X)

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

S(ok(X)) -> S(X)
S(mark(X)) -> S(X)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

S(x₁) -> S(x₁)
ok(x₁) -> ok(x₁)
mark(x₁) -> mark(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 10

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pair:

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pairs:

DIV(ok(X1), ok(X2)) -> DIV(X1, X2)
DIV(mark(X1), X2) -> DIV(X1, X2)

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

DIV(ok(X1), ok(X2)) -> DIV(X1, X2)
DIV(mark(X1), X2) -> DIV(X1, X2)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

DIV(x₁, x₂) -> DIV(x₁, x₂)
ok(x₁) -> ok(x₁)
mark(x₁) -> mark(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 11

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pair:

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Argument Filtering and Ordering

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pair:

MINUS(ok(X1), ok(X2)) -> MINUS(X1, X2)

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(ok(X1), ok(X2)) -> MINUS(X1, X2)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
ok(x₁) -> ok(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 12

             ↳Dependency Graph

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pair:

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Argument Filtering and Ordering

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pair:

GEQ(ok(X1), ok(X2)) -> GEQ(X1, X2)

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

GEQ(ok(X1), ok(X2)) -> GEQ(X1, X2)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

GEQ(x₁, x₂) -> GEQ(x₁, x₂)
ok(x₁) -> ok(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

           →DP Problem 13

             ↳Dependency Graph

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pair:

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳Argument Filtering and Ordering

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pairs:

ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(div(X1, X2)) -> ACTIVE(X1)
ACTIVE(s(X)) -> ACTIVE(X)

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

ACTIVE(if(X1, X2, X3)) -> ACTIVE(X1)
ACTIVE(div(X1, X2)) -> ACTIVE(X1)
ACTIVE(s(X)) -> ACTIVE(X)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

ACTIVE(x₁) -> ACTIVE(x₁)
s(x₁) -> s(x₁)
if(x₁, x₂, x₃) -> if(x₁, x₂, x₃)
div(x₁, x₂) -> div(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

           →DP Problem 14

             ↳Dependency Graph

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining

Dependency Pair:

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳Argument Filtering and Ordering

       →DP Problem 8

         ↳Remaining

Dependency Pairs:

PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(div(X1, X2)) -> PROPER(X2)
PROPER(div(X1, X2)) -> PROPER(X1)
PROPER(geq(X1, X2)) -> PROPER(X2)
PROPER(geq(X1, X2)) -> PROPER(X1)
PROPER(s(X)) -> PROPER(X)
PROPER(minus(X1, X2)) -> PROPER(X2)
PROPER(minus(X1, X2)) -> PROPER(X1)

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

PROPER(if(X1, X2, X3)) -> PROPER(X3)
PROPER(if(X1, X2, X3)) -> PROPER(X2)
PROPER(if(X1, X2, X3)) -> PROPER(X1)
PROPER(div(X1, X2)) -> PROPER(X2)
PROPER(div(X1, X2)) -> PROPER(X1)
PROPER(geq(X1, X2)) -> PROPER(X2)
PROPER(geq(X1, X2)) -> PROPER(X1)
PROPER(s(X)) -> PROPER(X)
PROPER(minus(X1, X2)) -> PROPER(X2)
PROPER(minus(X1, X2)) -> PROPER(X1)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

PROPER(x₁) -> PROPER(x₁)
div(x₁, x₂) -> div(x₁, x₂)
geq(x₁, x₂) -> geq(x₁, x₂)
minus(x₁, x₂) -> minus(x₁, x₂)
if(x₁, x₂, x₃) -> if(x₁, x₂, x₃)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

           →DP Problem 15

             ↳Dependency Graph

       →DP Problem 8

         ↳Remaining

Dependency Pair:

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

       →DP Problem 6

         ↳AFS

       →DP Problem 7

         ↳AFS

       →DP Problem 8

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

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

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))
s(ok(X)) -> ok(s(X))
div(mark(X1), X2) -> mark(div(X1, X2))
div(ok(X1), ok(X2)) -> ok(div(X1, X2))
if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3)) -> ok(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))
geq(ok(X1), ok(X2)) -> ok(geq(X1, X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:06 minutes