le(0,

le(s(

le(s(

pred(s(

minus(

minus(

gcd(0,

gcd(s(

gcd(s(

if

if

R

↳Dependency Pair Analysis

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

MINUS(x, s(y)) -> PRED(minus(x,y))

MINUS(x, s(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

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

→DP Problem 3

↳Remaining

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

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

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

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(x,_{1}x) -> LE(_{2}x,_{1}x)_{2}

s(x) -> s(_{1}x)_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳AFS

→DP Problem 3

↳Remaining

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and Ordering

→DP Problem 3

↳Remaining

**MINUS( x, s(y)) -> MINUS(x, y)**

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

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

The following dependency pair can be strictly oriented:

MINUS(x, s(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:

trivial

resulting in one new DP problem.

Used Argument Filtering System:

MINUS(x,_{1}x) -> MINUS(_{2}x,_{1}x)_{2}

s(x) -> s(_{1}x)_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Remaining

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳Remaining Obligation(s)

The following remains to be proven:

**IF _{GCD}(false, s(x), s(y)) -> GCD(minus(y, x), s(x))**

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

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

Duration:

0:02 minutes