3.40/1.60	WORST_CASE(NON_POLY, ?)
3.40/1.62	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.40/1.62	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.40/1.62	
3.40/1.62	
3.40/1.62	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.40/1.62	
3.40/1.62	(0) CpxTRS
3.40/1.62	(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
3.40/1.62	(2) TRS for Loop Detection
3.40/1.62	(3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
3.40/1.62	(4) BEST
3.40/1.62	    (5) proven lower bound
3.40/1.62	        (6) LowerBoundPropagationProof [FINISHED, 0 ms]
3.40/1.62	        (7) BOUNDS(n^1, INF)
3.40/1.62	    (8) TRS for Loop Detection
3.40/1.62	        (9) DecreasingLoopProof [FINISHED, 19 ms]
3.40/1.62	        (10) BOUNDS(EXP, INF)
3.40/1.62	
3.40/1.62	
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(0)
3.40/1.62	Obligation:
3.40/1.62	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.40/1.62	
3.40/1.62	
3.40/1.62	The TRS R consists of the following rules:
3.40/1.62	
3.40/1.62	   0(#) -> #
3.40/1.62	   +(x, #) -> x
3.40/1.62	   +(#, x) -> x
3.40/1.62	   +(0(x), 0(y)) -> 0(+(x, y))
3.40/1.62	   +(0(x), 1(y)) -> 1(+(x, y))
3.40/1.62	   +(1(x), 0(y)) -> 1(+(x, y))
3.40/1.62	   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
3.40/1.62	   *(#, x) -> #
3.40/1.62	   *(0(x), y) -> 0(*(x, y))
3.40/1.62	   *(1(x), y) -> +(0(*(x, y)), y)
3.40/1.62	   sum(nil) -> 0(#)
3.40/1.62	   sum(cons(x, l)) -> +(x, sum(l))
3.40/1.62	   prod(nil) -> 1(#)
3.40/1.62	   prod(cons(x, l)) -> *(x, prod(l))
3.40/1.62	
3.40/1.62	S is empty.
3.40/1.62	Rewrite Strategy: FULL
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
3.40/1.62	Transformed a relative TRS into a decreasing-loop problem.
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(2)
3.40/1.62	Obligation:
3.40/1.62	Analyzing the following TRS for decreasing loops:
3.40/1.62	
3.40/1.62	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.40/1.62	
3.40/1.62	
3.40/1.62	The TRS R consists of the following rules:
3.40/1.62	
3.40/1.62	   0(#) -> #
3.40/1.62	   +(x, #) -> x
3.40/1.62	   +(#, x) -> x
3.40/1.62	   +(0(x), 0(y)) -> 0(+(x, y))
3.40/1.62	   +(0(x), 1(y)) -> 1(+(x, y))
3.40/1.62	   +(1(x), 0(y)) -> 1(+(x, y))
3.40/1.62	   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
3.40/1.62	   *(#, x) -> #
3.40/1.62	   *(0(x), y) -> 0(*(x, y))
3.40/1.62	   *(1(x), y) -> +(0(*(x, y)), y)
3.40/1.62	   sum(nil) -> 0(#)
3.40/1.62	   sum(cons(x, l)) -> +(x, sum(l))
3.40/1.62	   prod(nil) -> 1(#)
3.40/1.62	   prod(cons(x, l)) -> *(x, prod(l))
3.40/1.62	
3.40/1.62	S is empty.
3.40/1.62	Rewrite Strategy: FULL
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(3) DecreasingLoopProof (LOWER BOUND(ID))
3.40/1.62	The following loop(s) give(s) rise to the lower bound Omega(n^1):
3.40/1.62	
3.40/1.62	The rewrite sequence
3.40/1.62	
3.40/1.62	+(1(x), 1(y)) ->^+ 0(+(+(x, y), 1(#)))
3.40/1.62	
3.40/1.62	gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
3.40/1.62	
3.40/1.62	The pumping substitution is [x / 1(x), y / 1(y)].
3.40/1.62	
3.40/1.62	The result substitution is [ ].
3.40/1.62	
3.40/1.62	
3.40/1.62	
3.40/1.62	
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(4)
3.40/1.62	Complex Obligation (BEST)
3.40/1.62	
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(5)
3.40/1.62	Obligation:
3.40/1.62	Proved the lower bound n^1 for the following obligation:
3.40/1.62	
3.40/1.62	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.40/1.62	
3.40/1.62	
3.40/1.62	The TRS R consists of the following rules:
3.40/1.62	
3.40/1.62	   0(#) -> #
3.40/1.62	   +(x, #) -> x
3.40/1.62	   +(#, x) -> x
3.40/1.62	   +(0(x), 0(y)) -> 0(+(x, y))
3.40/1.62	   +(0(x), 1(y)) -> 1(+(x, y))
3.40/1.62	   +(1(x), 0(y)) -> 1(+(x, y))
3.40/1.62	   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
3.40/1.62	   *(#, x) -> #
3.40/1.62	   *(0(x), y) -> 0(*(x, y))
3.40/1.62	   *(1(x), y) -> +(0(*(x, y)), y)
3.40/1.62	   sum(nil) -> 0(#)
3.40/1.62	   sum(cons(x, l)) -> +(x, sum(l))
3.40/1.62	   prod(nil) -> 1(#)
3.40/1.62	   prod(cons(x, l)) -> *(x, prod(l))
3.40/1.62	
3.40/1.62	S is empty.
3.40/1.62	Rewrite Strategy: FULL
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(6) LowerBoundPropagationProof (FINISHED)
3.40/1.62	Propagated lower bound.
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(7)
3.40/1.62	BOUNDS(n^1, INF)
3.40/1.62	
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(8)
3.40/1.62	Obligation:
3.40/1.62	Analyzing the following TRS for decreasing loops:
3.40/1.62	
3.40/1.62	The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF).
3.40/1.62	
3.40/1.62	
3.40/1.62	The TRS R consists of the following rules:
3.40/1.62	
3.40/1.62	   0(#) -> #
3.40/1.62	   +(x, #) -> x
3.40/1.62	   +(#, x) -> x
3.40/1.62	   +(0(x), 0(y)) -> 0(+(x, y))
3.40/1.62	   +(0(x), 1(y)) -> 1(+(x, y))
3.40/1.62	   +(1(x), 0(y)) -> 1(+(x, y))
3.40/1.62	   +(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
3.40/1.62	   *(#, x) -> #
3.40/1.62	   *(0(x), y) -> 0(*(x, y))
3.40/1.62	   *(1(x), y) -> +(0(*(x, y)), y)
3.40/1.62	   sum(nil) -> 0(#)
3.40/1.62	   sum(cons(x, l)) -> +(x, sum(l))
3.40/1.62	   prod(nil) -> 1(#)
3.40/1.62	   prod(cons(x, l)) -> *(x, prod(l))
3.40/1.62	
3.40/1.62	S is empty.
3.40/1.62	Rewrite Strategy: FULL
3.40/1.62	----------------------------------------
3.40/1.62	
3.40/1.62	(9) DecreasingLoopProof (FINISHED)
3.40/1.62	The following loop(s) give(s) rise to the lower bound EXP:
3.40/1.63	
3.40/1.63	The rewrite sequence
3.40/1.63	
3.40/1.63	prod(cons(1(x1_0), l)) ->^+ +(0(*(x1_0, prod(l))), prod(l))
3.40/1.63	
3.40/1.63	gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0,1].
3.40/1.63	
3.40/1.63	The pumping substitution is [l / cons(1(x1_0), l)].
3.40/1.63	
3.40/1.63	The result substitution is [ ].
3.40/1.63	
3.40/1.63	
3.40/1.63	
3.40/1.63	The rewrite sequence
3.40/1.63	
3.40/1.63	prod(cons(1(x1_0), l)) ->^+ +(0(*(x1_0, prod(l))), prod(l))
3.40/1.63	
3.40/1.63	gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
3.40/1.63	
3.40/1.63	The pumping substitution is [l / cons(1(x1_0), l)].
3.40/1.63	
3.40/1.63	The result substitution is [ ].
3.40/1.63	
3.40/1.63	
3.40/1.63	
3.40/1.63	
3.40/1.63	----------------------------------------
3.40/1.63	
3.40/1.63	(10)
3.40/1.63	BOUNDS(EXP, INF)
3.65/1.67	EOF