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))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Innermost 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)
GCD(s(x), s(y)) -> IF_GCD(le(y, x), s(x), s(y))
GCD(s(x), s(y)) -> LE(y, x)
IF_GCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
IF_GCD(true, s(x), s(y)) -> MINUS(x, y)
IF_GCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IF_GCD(false, s(x), s(y)) -> MINUS(y, x)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Remaining

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))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

LE(x₁, x₂) -> LE(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Remaining

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))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(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

         ↳Remaining

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))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

MINUS(x₁, x₂) -> x₁
s(x₁) -> s(x₁)
IF_MINUS(x₁, x₂, x₃) -> x₂

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Remaining

Dependency Pair:

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))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(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

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

IF_GCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IF_GCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
GCD(s(x), s(y)) -> IF_GCD(le(y, x), s(x), 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))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes