TRS Equat 89423 pair #381732808

details

property	value
status	complete
benchmark	AC11.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n049.star.cs.uiowa.edu
space	AProVE_AC_04

run statistics

property	value
solver	muterm 5.18
configuration	default
runtime (wallclock)	0.15283703804 seconds
cpu usage	0.149016714
max memory	4112384.0

stage attributes

key	value
output-size	7785
starexec-result	YES

output

/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: 

(THEORY 
(AC f))
(RULES
f(g(f(h(a),a)),a) -> f(h(a),f(a,a))
f(g(h(a)),f(b,f(b,b))) -> f(g(f(h(a),a)),a)
f(h(a),a) -> f(h(a),b)
f(h(a),g(a)) -> f(g(h(a)),a)
)

Problem 1: 

Reduction Order Processor:
-> Rules:
 f(g(f(h(a),a)),a) -> f(h(a),f(a,a))
 f(g(h(a)),f(b,f(b,b))) -> f(g(f(h(a),a)),a)
 f(h(a),a) -> f(h(a),b)
 f(h(a),g(a)) -> f(g(h(a)),a)
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[f](X1,X2) = X1 + X2 + 2
[a] = 2
[b] = 2
[g](X) = X + 2
[h](X) = 2.X + 2

Problem 1: 

Reduction Order Processor:
-> Rules:
 f(g(h(a)),f(b,f(b,b))) -> f(g(f(h(a),a)),a)
 f(h(a),a) -> f(h(a),b)
 f(h(a),g(a)) -> f(g(h(a)),a)
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[f](X1,X2) = X1 + X2 + 2
[a] = 1
[b] = 1
[g](X) = X + 2
[h](X) = 2.X + 2

Problem 1: 

Reduction Order Processor:
-> Rules:
 f(h(a),a) -> f(h(a),b)
 f(h(a),g(a)) -> f(g(h(a)),a)
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[f](X1,X2) = X1 + X2 + 2
[a] = 1
[b] = 0
[g](X) = 2.X + 2
[h](X) = X

Problem 1: 

Dependency Pairs Processor:
-> FAxioms:
 F(f(x0,x1),x2) = F(x0,f(x1,x2))
 F(x0,x1) = F(x1,x0)
-> Pairs:
 F(f(h(a),g(a)),x0) -> F(f(g(h(a)),a),x0)
 F(f(h(a),g(a)),x0) -> F(g(h(a)),a)
 F(h(a),g(a)) -> F(g(h(a)),a)
-> EAxioms:
 f(f(x0,x1),x2) = f(x0,f(x1,x2))

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