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))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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)
MOD(s(x), s(y)) -> IF_MOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) -> LE(y, x)
IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
IF_MOD(true, s(x), s(y)) -> MINUS(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳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))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

         ↳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))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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

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))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

{MINUS, IF_MINUS}

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

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))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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

Dependency Pairs:

IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
MOD(s(x), s(y)) -> IF_MOD(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))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

The following dependency pair can be strictly oriented:

IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))

The following usable rules w.r.t. to the AFS 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))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

{MOD, IF_MOD} > 0
if_minus > 0
true > 0
minus > 0
s > 0
le > 0

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

MOD(x₁, x₂) -> MOD(x₁, x₂)
IF_MOD(x₁, x₂, x₃) -> IF_MOD(x₂, x₃)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> x₁
if_minus(x₁, x₂, x₃) -> x₂

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

Dependency Pair:

MOD(s(x), s(y)) -> IF_MOD(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))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:02 minutes