/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

Problem 1: 

(VAR x y)
(RULES
and(not(x),x) -> 0
and(0,x) -> 0
and(1,x) -> x
and(x,not(x)) -> 0
and(x,0) -> 0
and(x,1) -> x
f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
implies(x,y) -> 0 | x -> 1, y -> 0
implies(x,y) -> 1 | not(x) -> 1
implies(x,y) -> 1 | y -> 1
not(0) -> 1
not(1) -> 0
or(not(x),x) -> 1
or(0,x) -> x
or(1,x) -> 1
or(x,not(x)) -> 1
or(x,0) -> x
or(x,1) -> 1
)

Problem 1: 
Valid CTRS Processor:
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
 implies(x,y) -> 0 | x -> 1, y -> 0
 implies(x,y) -> 1 | not(x) -> 1
 implies(x,y) -> 1 | y -> 1
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
-> The system is a deterministic 3-CTRS.

Problem 1: 

Dependency Pairs Processor:

Conditional Termination Problem 1:
-> Pairs:
 F(x) -> F(0) | implies(implies(x,implies(x,0)),0) -> 1
-> QPairs:
 Empty
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
 implies(x,y) -> 0 | x -> 1, y -> 0
 implies(x,y) -> 1 | not(x) -> 1
 implies(x,y) -> 1 | y -> 1
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1

Conditional Termination Problem 2:
-> Pairs:
 F(x) -> IMPLIES(implies(x,implies(x,0)),0)
 F(x) -> IMPLIES(x,implies(x,0))
 F(x) -> IMPLIES(x,0)
 IMPLIES(x,y) -> NOT(x)
-> QPairs:
 F(x) -> F(0) | implies(implies(x,implies(x,0)),0) -> 1
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
 implies(x,y) -> 0 | x -> 1, y -> 0
 implies(x,y) -> 1 | not(x) -> 1
 implies(x,y) -> 1 | y -> 1
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1


The problem is decomposed in 2 subproblems.

Problem 1.1: 

SCC Processor:
-> Pairs:
 F(x) -> F(0) | implies(implies(x,implies(x,0)),0) -> 1
-> QPairs:
 Empty
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
 implies(x,y) -> 0 | x -> 1, y -> 0
 implies(x,y) -> 1 | not(x) -> 1
 implies(x,y) -> 1 | y -> 1
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(x) -> F(0) | implies(implies(x,implies(x,0)),0) -> 1
->->-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
 implies(x,y) -> 0 | x -> 1, y -> 0
 implies(x,y) -> 1 | not(x) -> 1
 implies(x,y) -> 1 | y -> 1
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1

Problem 1.1: 

Ohlebusch Transformation Processor:
-> Pairs:
 F(x) -> F(0) | implies(implies(x,implies(x,0)),0) -> 1
-> QPairs:
 Empty
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
 implies(x,y) -> 0 | x -> 1, y -> 0
 implies(x,y) -> 1 | not(x) -> 1
 implies(x,y) -> 1 | y -> 1
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
-> New Pairs:
 F(x) -> U15(implies(implies(x,implies(x,0)),0),x)
 U15(1,x) -> F(0)
-> New Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> U10(implies(implies(x,implies(x,0)),0),x)
 implies(x,y) -> U11(x,x,y)
 implies(x,y) -> U13(not(x),x,y)
 implies(x,y) -> U14(y,x,y)
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
 U10(1,x) -> f(0)
 U11(1,x,y) -> U12(y,x,y)
 U12(0,x,y) -> 0
 U13(1,x,y) -> 1
 U14(1,x,y) -> 1

Problem 1.1: 

SCC Processor:
-> Pairs:
 F(x) -> U15(implies(implies(x,implies(x,0)),0),x)
 U15(1,x) -> F(0)
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> U10(implies(implies(x,implies(x,0)),0),x)
 implies(x,y) -> U11(x,x,y)
 implies(x,y) -> U13(not(x),x,y)
 implies(x,y) -> U14(y,x,y)
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
 U10(1,x) -> f(0)
 U11(1,x,y) -> U12(y,x,y)
 U12(0,x,y) -> 0
 U13(1,x,y) -> 1
 U14(1,x,y) -> 1
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(x) -> U15(implies(implies(x,implies(x,0)),0),x)
 U15(1,x) -> F(0)
->->-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> U10(implies(implies(x,implies(x,0)),0),x)
 implies(x,y) -> U11(x,x,y)
 implies(x,y) -> U13(not(x),x,y)
 implies(x,y) -> U14(y,x,y)
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
 U10(1,x) -> f(0)
 U11(1,x,y) -> U12(y,x,y)
 U12(0,x,y) -> 0
 U13(1,x,y) -> 1
 U14(1,x,y) -> 1

Problem 1.1: 

Reduction Pair Processor:
-> Pairs:
 F(x) -> U15(implies(implies(x,implies(x,0)),0),x)
 U15(1,x) -> F(0)
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> U10(implies(implies(x,implies(x,0)),0),x)
 implies(x,y) -> U11(x,x,y)
 implies(x,y) -> U13(not(x),x,y)
 implies(x,y) -> U14(y,x,y)
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
 U10(1,x) -> f(0)
 U11(1,x,y) -> U12(y,x,y)
 U12(0,x,y) -> 0
 U13(1,x,y) -> 1
 U14(1,x,y) -> 1
-> Usable rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> U10(implies(implies(x,implies(x,0)),0),x)
 implies(x,y) -> U11(x,x,y)
 implies(x,y) -> U13(not(x),x,y)
 implies(x,y) -> U14(y,x,y)
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
 U10(1,x) -> f(0)
 U11(1,x,y) -> U12(y,x,y)
 U12(0,x,y) -> 0
 U13(1,x,y) -> 1
 U14(1,x,y) -> 1
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 2
->Bound:
 1
->Interpretation:
 
[and](X1,X2) = [1 1;0 1].X1 + [1 1;0 1].X2 + [0;1]
[f](X) = [0 1;1 1].X + [1;1]
[implies](X1,X2) = [0 1;1 0].X1 + [0 0;0 1].X2
[not](X) = [1 0;1 0].X + [1;0]
[or](X1,X2) = [1 0;1 1].X1 + [1 0;0 1].X2 + [1;1]
[0] = [1;0]
[1] = [0;1]
[F](X) = [0 1;0 1].X + [1;1]
[U10](X1,X2) = [0 0;0 1].X1 + [0 1;1 0].X2 + [1;1]
[U11](X1,X2,X3) = [0 1;0 0].X1 + [0 0;1 0].X2 + [0 0;0 1].X3
[U12](X1,X2,X3) = [0 0;0 1].X1 + [0 0;1 0].X2 + [1;0]
[U13](X1,X2,X3) = [0 0;0 1].X1 + [0 1;0 0].X2 + [0 0;0 1].X3
[U14](X1,X2,X3) = [0 0;0 1].X1 + [0 0;1 0].X2
[U15](X1,X2) = [0 1;0 0].X1 + [0 0;0 1].X2 + [0;1]

Problem 1.1: 

SCC Processor:
-> Pairs:
 U15(1,x) -> F(0)
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> U10(implies(implies(x,implies(x,0)),0),x)
 implies(x,y) -> U11(x,x,y)
 implies(x,y) -> U13(not(x),x,y)
 implies(x,y) -> U14(y,x,y)
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
 U10(1,x) -> f(0)
 U11(1,x,y) -> U12(y,x,y)
 U12(0,x,y) -> 0
 U13(1,x,y) -> 1
 U14(1,x,y) -> 1
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

SCC Processor:
-> Pairs:
 F(x) -> IMPLIES(implies(x,implies(x,0)),0)
 F(x) -> IMPLIES(x,implies(x,0))
 F(x) -> IMPLIES(x,0)
 IMPLIES(x,y) -> NOT(x)
-> QPairs:
 F(x) -> F(0) | implies(implies(x,implies(x,0)),0) -> 1
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
 implies(x,y) -> 0 | x -> 1, y -> 0
 implies(x,y) -> 1 | not(x) -> 1
 implies(x,y) -> 1 | y -> 1
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.