/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 v_NonEmpty:S x:S y:S)
(RULES
and(not(x:S),x:S) -> 0
and(0,x:S) -> 0
and(1,x:S) -> x:S
and(x:S,not(x:S)) -> 0
and(x:S,0) -> 0
and(x:S,1) -> x:S
f(x:S) -> f(0) | implies(implies(x:S,implies(x:S,0)),0) ->* 1
implies(x:S,y:S) -> 0 | x:S ->* 1, y:S ->* 0
implies(x:S,y:S) -> 1 | not(x:S) ->* 1
implies(x:S,y:S) -> 1 | y:S ->* 1
not(0) -> 1
not(1) -> 0
or(not(x:S),x:S) -> 1
or(0,x:S) -> x:S
or(1,x:S) -> 1
or(x:S,not(x:S)) -> 1
or(x:S,0) -> x:S
or(x:S,1) -> 1
)

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

Problem 1: 

Dependency Pairs Processor:

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

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


The problem is decomposed in 2 subproblems.

Problem 1.1: 

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

Problem 1.1: 

Simplifying Unifiable Condition Processor:
-> Pairs:
 F(x:S) -> F(0) | implies(implies(x:S,implies(x:S,0)),0) ->* 1
-> QPairs:
 Empty
-> Rules:
 and(not(x:S),x:S) -> 0
 and(0,x:S) -> 0
 and(1,x:S) -> x:S
 and(x:S,not(x:S)) -> 0
 and(x:S,0) -> 0
 and(x:S,1) -> x:S
 f(x:S) -> f(0) | implies(implies(x:S,implies(x:S,0)),0) ->* 1
 implies(x:S,y:S) -> 0 | x:S ->* 1, y:S ->* 0
 implies(x:S,y:S) -> 1 | not(x:S) ->* 1
 implies(x:S,y:S) -> 1 | y:S ->* 1
 not(0) -> 1
 not(1) -> 0
 or(not(x:S),x:S) -> 1
 or(0,x:S) -> x:S
 or(1,x:S) -> 1
 or(x:S,not(x:S)) -> 1
 or(x:S,0) -> x:S
 or(x:S,1) -> 1
->Transformed Pairs/Rules:
->->Original Pair/Rule:
 implies(x:S,y:S) -> 0 | x:S ->* 1, y:S ->* 0
->-> Transformed Pair/Rule:
 implies(1,0) -> 0
->->Original Pair/Rule:
 implies(x:S,y:S) -> 1 | y:S ->* 1
->-> Transformed Pair/Rule:
 implies(x:S,1) -> 1

Problem 1.1: 

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

Problem 1.1: 

Ohlebusch Transformation Processor:
-> Pairs:
 F(x:S) -> F(0) | implies(implies(x:S,implies(x:S,0)),0) ->* 1
-> QPairs:
 Empty
-> Rules:
 and(not(x:S),x:S) -> 0
 and(0,x:S) -> 0
 and(1,x:S) -> x:S
 and(x:S,not(x:S)) -> 0
 and(x:S,0) -> 0
 and(x:S,1) -> x:S
 f(x:S) -> f(0) | implies(implies(x:S,implies(x:S,0)),0) ->* 1
 implies(1,0) -> 0
 implies(x:S,1) -> 1
 implies(x:S,y:S) -> 1 | not(x:S) ->* 1
 not(0) -> 1
 not(1) -> 0
 or(not(x:S),x:S) -> 1
 or(0,x:S) -> x:S
 or(1,x:S) -> 1
 or(x:S,not(x:S)) -> 1
 or(x:S,0) -> x:S
 or(x:S,1) -> 1
-> New Pairs:
 F(x:S) -> U17(implies(implies(x:S,implies(x:S,0)),0),x:S)
 U17(1,x:S) -> F(0)
-> New Rules:
 and(not(x:S),x:S) -> 0
 and(0,x:S) -> 0
 and(1,x:S) -> x:S
 and(x:S,not(x:S)) -> 0
 and(x:S,0) -> 0
 and(x:S,1) -> x:S
 f(x:S) -> U15(implies(implies(x:S,implies(x:S,0)),0),x:S)
 implies(1,0) -> 0
 implies(x:S,1) -> 1
 implies(x:S,y:S) -> U16(not(x:S),x:S,y:S)
 not(0) -> 1
 not(1) -> 0
 or(not(x:S),x:S) -> 1
 or(0,x:S) -> x:S
 or(1,x:S) -> 1
 or(x:S,not(x:S)) -> 1
 or(x:S,0) -> x:S
 or(x:S,1) -> 1
 U15(1,x:S) -> f(0)
 U16(1,x:S,y:S) -> 1

Problem 1.1: 

SCC Processor:
-> Pairs:
 F(x:S) -> U17(implies(implies(x:S,implies(x:S,0)),0),x:S)
 U17(1,x:S) -> F(0)
-> Rules:
 and(not(x:S),x:S) -> 0
 and(0,x:S) -> 0
 and(1,x:S) -> x:S
 and(x:S,not(x:S)) -> 0
 and(x:S,0) -> 0
 and(x:S,1) -> x:S
 f(x:S) -> U15(implies(implies(x:S,implies(x:S,0)),0),x:S)
 implies(1,0) -> 0
 implies(x:S,1) -> 1
 implies(x:S,y:S) -> U16(not(x:S),x:S,y:S)
 not(0) -> 1
 not(1) -> 0
 or(not(x:S),x:S) -> 1
 or(0,x:S) -> x:S
 or(1,x:S) -> 1
 or(x:S,not(x:S)) -> 1
 or(x:S,0) -> x:S
 or(x:S,1) -> 1
 U15(1,x:S) -> f(0)
 U16(1,x:S,y:S) -> 1
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(x:S) -> U17(implies(implies(x:S,implies(x:S,0)),0),x:S)
 U17(1,x:S) -> F(0)
->->-> Rules:
 and(not(x:S),x:S) -> 0
 and(0,x:S) -> 0
 and(1,x:S) -> x:S
 and(x:S,not(x:S)) -> 0
 and(x:S,0) -> 0
 and(x:S,1) -> x:S
 f(x:S) -> U15(implies(implies(x:S,implies(x:S,0)),0),x:S)
 implies(1,0) -> 0
 implies(x:S,1) -> 1
 implies(x:S,y:S) -> U16(not(x:S),x:S,y:S)
 not(0) -> 1
 not(1) -> 0
 or(not(x:S),x:S) -> 1
 or(0,x:S) -> x:S
 or(1,x:S) -> 1
 or(x:S,not(x:S)) -> 1
 or(x:S,0) -> x:S
 or(x:S,1) -> 1
 U15(1,x:S) -> f(0)
 U16(1,x:S,y:S) -> 1

Problem 1.1: 

Reduction Pair Processor:
-> Pairs:
 F(x:S) -> U17(implies(implies(x:S,implies(x:S,0)),0),x:S)
 U17(1,x:S) -> F(0)
-> Rules:
 and(not(x:S),x:S) -> 0
 and(0,x:S) -> 0
 and(1,x:S) -> x:S
 and(x:S,not(x:S)) -> 0
 and(x:S,0) -> 0
 and(x:S,1) -> x:S
 f(x:S) -> U15(implies(implies(x:S,implies(x:S,0)),0),x:S)
 implies(1,0) -> 0
 implies(x:S,1) -> 1
 implies(x:S,y:S) -> U16(not(x:S),x:S,y:S)
 not(0) -> 1
 not(1) -> 0
 or(not(x:S),x:S) -> 1
 or(0,x:S) -> x:S
 or(1,x:S) -> 1
 or(x:S,not(x:S)) -> 1
 or(x:S,0) -> x:S
 or(x:S,1) -> 1
 U15(1,x:S) -> f(0)
 U16(1,x:S,y:S) -> 1
-> Usable rules:
 and(not(x:S),x:S) -> 0
 and(0,x:S) -> 0
 and(1,x:S) -> x:S
 and(x:S,not(x:S)) -> 0
 and(x:S,0) -> 0
 and(x:S,1) -> x:S
 f(x:S) -> U15(implies(implies(x:S,implies(x:S,0)),0),x:S)
 implies(1,0) -> 0
 implies(x:S,1) -> 1
 implies(x:S,y:S) -> U16(not(x:S),x:S,y:S)
 not(0) -> 1
 not(1) -> 0
 or(not(x:S),x:S) -> 1
 or(0,x:S) -> x:S
 or(1,x:S) -> 1
 or(x:S,not(x:S)) -> 1
 or(x:S,0) -> x:S
 or(x:S,1) -> 1
 U15(1,x:S) -> f(0)
 U16(1,x:S,y:S) -> 1
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 2
->Bound:
 1
->Interpretation:
 
[and](X1,X2) = [1 0;1 1].X1 + [1 0;0 1].X2 + [1;0]
[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 1;1 1].X2
[0] = [1;0]
[1] = [0;1]
[F](X) = [0 1;0 1].X + [1;1]
[U15](X1,X2) = [0 0;0 1].X1 + [0 0;1 0].X2 + [1;1]
[U16](X1,X2,X3) = [0 0;0 1].X1 + [0 1;0 0].X2
[U17](X1,X2) = [0 1;0 0].X1 + [0 0;0 1].X2 + [0;1]

Problem 1.1: 

SCC Processor:
-> Pairs:
 U17(1,x:S) -> F(0)
-> Rules:
 and(not(x:S),x:S) -> 0
 and(0,x:S) -> 0
 and(1,x:S) -> x:S
 and(x:S,not(x:S)) -> 0
 and(x:S,0) -> 0
 and(x:S,1) -> x:S
 f(x:S) -> U15(implies(implies(x:S,implies(x:S,0)),0),x:S)
 implies(1,0) -> 0
 implies(x:S,1) -> 1
 implies(x:S,y:S) -> U16(not(x:S),x:S,y:S)
 not(0) -> 1
 not(1) -> 0
 or(not(x:S),x:S) -> 1
 or(0,x:S) -> x:S
 or(1,x:S) -> 1
 or(x:S,not(x:S)) -> 1
 or(x:S,0) -> x:S
 or(x:S,1) -> 1
 U15(1,x:S) -> f(0)
 U16(1,x:S,y:S) -> 1
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

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

The problem is finite.