Termination Proof using AProVE

Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(s(x), y) -> IF_MINUS(le(s(x), y), s(x), y)
MINUS(s(x), y) -> LE(s(x), y)
IF_MINUS(false, s(x), y) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))
LOG(s(s(x))) -> QUOT(x, s(s(0)))

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The following dependency pair can be strictly oriented:

LE(s(x), s(y)) -> LE(x, y)

The following rules can be oriented:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(LE(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(false) = 0

POL(log(x₁)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(le(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.
Used Argument Filtering System:

LE(x₁, x₂) -> LE(x₁, x₂)
s(x₁) -> s(x₁)
le(x₁, x₂) -> le(x₁, x₂)
minus(x₁, x₂) -> x₁
if_minus(x₁, x₂, x₃) -> x₂
quot(x₁, x₂) -> x₁
log(x₁) -> log(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

Dependency Pairs:

IF_MINUS(false, s(x), y) -> MINUS(x, y)
MINUS(s(x), y) -> IF_MINUS(le(s(x), y), s(x), y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The following dependency pair can be strictly oriented:

MINUS(s(x), y) -> IF_MINUS(le(s(x), y), s(x), y)

The following rules can be oriented:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IF_MINUS(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(0) = 0

POL(false) = 0

POL(log(x₁)) = x₁

POL(MINUS(x₁, x₂)) = 1 + x₁ + x₂

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(le) = 0

resulting in one new DP problem.
Used Argument Filtering System:

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
IF_MINUS(x₁, x₂, x₃) -> IF_MINUS(x₁, x₂, x₃)
s(x₁) -> s(x₁)
le(x₁, x₂) -> le
minus(x₁, x₂) -> x₁
if_minus(x₁, x₂, x₃) -> x₂
quot(x₁, x₂) -> x₁
log(x₁) -> log(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

IF_MINUS(false, s(x), y) -> MINUS(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The following dependency pair can be strictly oriented:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

The following rules can be oriented:

minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(false) = 0

POL(log(x₁)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(le(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.
Used Argument Filtering System:

QUOT(x₁, x₂) -> QUOT(x₁, x₂)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> x₁
if_minus(x₁, x₂, x₃) -> x₂
le(x₁, x₂) -> le(x₁, x₂)
quot(x₁, x₂) -> x₁
log(x₁) -> log(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

Dependency Pair:

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The following dependency pair can be strictly oriented:

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

The following rules can be oriented:

quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(false) = 0

POL(log(x₁)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(le(x₁, x₂)) = x₁ + x₂

POL(LOG(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

LOG(x₁) -> LOG(x₁)
s(x₁) -> s(x₁)
quot(x₁, x₂) -> x₁
minus(x₁, x₂) -> x₁
if_minus(x₁, x₂, x₃) -> x₂
le(x₁, x₂) -> le(x₁, x₂)
log(x₁) -> log(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 8

             ↳Dependency Graph

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes

POL(LE(x₁, x₂))	= 1 + x₁ + x₂
POL(0)	= 0
POL(false)	= 0
POL(log(x₁))	= x₁
POL(true)	= 0
POL(s(x₁))	= 1 + x₁
POL(le(x₁, x₂))	= x₁ + x₂

POL(IF_MINUS(x₁, x₂, x₃))	= x₁ + x₂ + x₃
POL(0)	= 0
POL(false)	= 0
POL(log(x₁))	= x₁
POL(MINUS(x₁, x₂))	= 1 + x₁ + x₂
POL(true)	= 0
POL(s(x₁))	= 1 + x₁
POL(le)	= 0

POL(QUOT(x₁, x₂))	= 1 + x₁ + x₂
POL(0)	= 0
POL(false)	= 0
POL(log(x₁))	= x₁
POL(true)	= 0
POL(s(x₁))	= 1 + x₁
POL(le(x₁, x₂))	= x₁ + x₂