Runtime_Complexity_Innermost_Rewriting 2019-04-01 06.40 pair #433313558

details

property	value
status	complete
benchmark	#4.32.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n081.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	5.00571 seconds
cpu usage	15.9818
user time	15.0048
system time	0.976983
max virtual memory	1.8809544E7
max residence set size	3101008.0

stage attributes

key	value
starexec-result	WORST_CASE(Omega(n^1), O(n^1))

output

15.85/4.92	WORST_CASE(Omega(n^1), O(n^1))
15.85/4.92	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
15.85/4.92	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
15.85/4.92	
15.85/4.92	
15.85/4.92	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).
15.85/4.92	
15.85/4.92	(0) CpxTRS
15.85/4.92	(1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
15.85/4.92	(2) CdtProblem
15.85/4.92	    (3) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 79 ms]
15.85/4.92	    (4) CdtProblem
15.85/4.92	    (5) CdtInstantiationProof [BOTH BOUNDS(ID, ID), 0 ms]
15.85/4.92	    (6) CdtProblem
15.85/4.92	    (7) CdtRhsSimplificationProcessorProof [BOTH BOUNDS(ID, ID), 0 ms]
15.85/4.92	    (8) CdtProblem
15.85/4.92	    (9) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
15.85/4.92	    (10) CdtProblem
15.85/4.92	    (11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 0 ms]
15.85/4.92	    (12) CdtProblem
15.85/4.92	    (13) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
15.85/4.92	    (14) BOUNDS(1, 1)
15.85/4.92	(15) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
15.85/4.92	(16) TRS for Loop Detection
15.85/4.92	    (17) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
15.85/4.92	    (18) BEST
15.85/4.92	        (19) proven lower bound
15.85/4.92	            (20) LowerBoundPropagationProof [FINISHED, 0 ms]
15.85/4.92	            (21) BOUNDS(n^1, INF)
15.85/4.92	        (22) TRS for Loop Detection
15.85/4.92	
15.85/4.92	
15.85/4.92	----------------------------------------
15.85/4.92	
15.85/4.92	(0)
15.85/4.92	Obligation:
15.85/4.92	The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).
15.85/4.92	
15.85/4.92	
15.85/4.92	The TRS R consists of the following rules:
15.85/4.92	
15.85/4.92	   f(x, c(y)) -> f(x, s(f(y, y)))
15.85/4.92	   f(s(x), s(y)) -> f(x, s(c(s(y))))
15.85/4.92	
15.85/4.92	S is empty.
15.85/4.92	Rewrite Strategy: INNERMOST
15.85/4.92	----------------------------------------
15.85/4.92	
15.85/4.92	(1) CpxTrsToCdtProof (UPPER BOUND(ID))
15.85/4.92	Converted Cpx (relative) TRS to CDT
15.85/4.92	----------------------------------------
15.85/4.92	
15.85/4.92	(2)
15.85/4.92	Obligation:
15.85/4.92	Complexity Dependency Tuples Problem
15.85/4.92	
15.85/4.92	Rules:
15.85/4.92	   f(z0, c(z1)) -> f(z0, s(f(z1, z1)))
15.85/4.92	   f(s(z0), s(z1)) -> f(z0, s(c(s(z1))))
15.85/4.92	Tuples:
15.85/4.92	   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
15.85/4.92	   F(s(z0), s(z1)) -> c2(F(z0, s(c(s(z1)))))
15.85/4.92	S tuples:
15.85/4.92	   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
15.85/4.92	   F(s(z0), s(z1)) -> c2(F(z0, s(c(s(z1)))))
15.85/4.92	K tuples:none
15.85/4.92	Defined Rule Symbols:   f_2
15.85/4.92	
15.85/4.92	Defined Pair Symbols:   F_2
15.85/4.92	
15.85/4.92	Compound Symbols:   c1_2, c2_1
15.85/4.92	
15.85/4.92	
15.85/4.92	----------------------------------------
15.85/4.92	
15.85/4.92	(3) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
15.85/4.92	Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
15.85/4.92	   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
15.85/4.92	We considered the (Usable) Rules:none
15.85/4.92	And the Tuples:
15.85/4.92	   F(z0, c(z1)) -> c1(F(z0, s(f(z1, z1))), F(z1, z1))
15.85/4.92	   F(s(z0), s(z1)) -> c2(F(z0, s(c(s(z1)))))
15.85/4.92	The order we found is given by the following interpretation:
15.85/4.92	
15.85/4.92	Polynomial interpretation :
15.85/4.92	
15.85/4.92	   POL(F(x_1, x_2)) = x_2
15.85/4.92	   POL(c(x_1)) = [1] + x_1
15.85/4.92	   POL(c1(x_1, x_2)) = x_1 + x_2
15.85/4.92	   POL(c2(x_1)) = x_1
15.85/4.92	   POL(f(x_1, x_2)) = [1] + x_2
15.85/4.92	   POL(s(x_1)) = 0
15.85/4.92	
15.85/4.92	----------------------------------------
15.85/4.92	
15.85/4.92	(4)
15.85/4.92	Obligation:
15.85/4.92	Complexity Dependency Tuples Problem
15.85/4.92	
15.85/4.92	Rules:

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