-(

-(s(

min(

min(0,

min(s(

twice(0) -> 0

twice(s(

f(s(

f(s(

R

↳Dependency Pair Analysis

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

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

TWICE(s(x)) -> TWICE(x)

F(s(x), s(y)) -> F(-(y, min(x,y)), s(twice(min(x,y))))

F(s(x), s(y)) -> -'(y, min(x,y))

F(s(x), s(y)) -> MIN(x,y)

F(s(x), s(y)) -> TWICE(min(x,y))

F(s(x), s(y)) -> F(-(x, min(x,y)), s(twice(min(x,y))))

F(s(x), s(y)) -> -'(x, min(x,y))

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Remaining

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

-(x, 0) ->x

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

min(x, 0) -> 0

min(0,y) -> 0

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

twice(0) -> 0

twice(s(x)) -> s(s(twice(x)))

f(s(x), s(y)) -> f(-(y, min(x,y)), s(twice(min(x,y))))

f(s(x), s(y)) -> f(-(x, min(x,y)), s(twice(min(x,y))))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

-(x, 0) ->x

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

min(x, 0) -> 0

min(0,y) -> 0

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

twice(0) -> 0

twice(s(x)) -> s(s(twice(x)))

f(s(x), s(y)) -> f(-(y, min(x,y)), s(twice(min(x,y))))

f(s(x), s(y)) -> f(-(x, min(x,y)), s(twice(min(x,y))))

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

twice > s

min > s

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

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

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

f(x,_{1}x) -> f_{2}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 5

↳Dependency Graph

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Remaining

-(x, 0) ->x

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

min(x, 0) -> 0

min(0,y) -> 0

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

twice(0) -> 0

twice(s(x)) -> s(s(twice(x)))

f(s(x), s(y)) -> f(-(y, min(x,y)), s(twice(min(x,y))))

f(s(x), s(y)) -> f(-(x, min(x,y)), s(twice(min(x,y))))

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

↳AFS

→DP Problem 4

↳Remaining

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

-(x, 0) ->x

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

min(x, 0) -> 0

min(0,y) -> 0

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

twice(0) -> 0

twice(s(x)) -> s(s(twice(x)))

f(s(x), s(y)) -> f(-(y, min(x,y)), s(twice(min(x,y))))

f(s(x), s(y)) -> f(-(x, min(x,y)), s(twice(min(x,y))))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

-(x, 0) ->x

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

min(x, 0) -> 0

min(0,y) -> 0

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

twice(0) -> 0

twice(s(x)) -> s(s(twice(x)))

f(s(x), s(y)) -> f(-(y, min(x,y)), s(twice(min(x,y))))

f(s(x), s(y)) -> f(-(x, min(x,y)), s(twice(min(x,y))))

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

twice > s

min > s

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

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

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

f(x,_{1}x) -> f_{2}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 6

↳Dependency Graph

→DP Problem 3

↳AFS

→DP Problem 4

↳Remaining

-(x, 0) ->x

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

min(x, 0) -> 0

min(0,y) -> 0

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

twice(0) -> 0

twice(s(x)) -> s(s(twice(x)))

f(s(x), s(y)) -> f(-(y, min(x,y)), s(twice(min(x,y))))

f(s(x), s(y)) -> f(-(x, min(x,y)), s(twice(min(x,y))))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳Argument Filtering and Ordering

→DP Problem 4

↳Remaining

**TWICE(s( x)) -> TWICE(x)**

-(x, 0) ->x

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

min(x, 0) -> 0

min(0,y) -> 0

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

twice(0) -> 0

twice(s(x)) -> s(s(twice(x)))

f(s(x), s(y)) -> f(-(y, min(x,y)), s(twice(min(x,y))))

f(s(x), s(y)) -> f(-(x, min(x,y)), s(twice(min(x,y))))

The following dependency pair can be strictly oriented:

TWICE(s(x)) -> TWICE(x)

The following rules can be oriented:

-(x, 0) ->x

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

min(x, 0) -> 0

min(0,y) -> 0

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

twice(0) -> 0

twice(s(x)) -> s(s(twice(x)))

f(s(x), s(y)) -> f(-(y, min(x,y)), s(twice(min(x,y))))

f(s(x), s(y)) -> f(-(x, min(x,y)), s(twice(min(x,y))))

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

twice > s

min > s

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

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

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

f(x,_{1}x) -> f_{2}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 7

↳Dependency Graph

→DP Problem 4

↳Remaining

-(x, 0) ->x

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

min(x, 0) -> 0

min(0,y) -> 0

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

twice(0) -> 0

twice(s(x)) -> s(s(twice(x)))

f(s(x), s(y)) -> f(-(y, min(x,y)), s(twice(min(x,y))))

f(s(x), s(y)) -> f(-(x, min(x,y)), s(twice(min(x,y))))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Remaining Obligation(s)

The following remains to be proven:

**F(s( x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))**

-(x, 0) ->x

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

min(x, 0) -> 0

min(0,y) -> 0

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

twice(0) -> 0

twice(s(x)) -> s(s(twice(x)))

f(s(x), s(y)) -> f(-(y, min(x,y)), s(twice(min(x,y))))

f(s(x), s(y)) -> f(-(x, min(x,y)), s(twice(min(x,y))))

Duration:

0:08 minutes