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


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) Strip Symbols Proof [SOUND, 0 ms]
(4) QTRS
(5) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms]
(6) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a(b(a(a(a(a(x1)))))) -> a(a(a(a(a(b(a(a(b(a(b(a(b(x1)))))))))))))

Q is empty.

----------------------------------------

(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a(a(a(a(b(a(x1)))))) -> b(a(b(a(b(a(a(b(a(a(a(a(a(x1)))))))))))))

Q is empty.

----------------------------------------

(3) Strip Symbols Proof (SOUND)
We were given the following TRS:

   a(a(a(a(b(a(x1)))))) -> b(a(b(a(b(a(a(b(a(a(a(a(a(x1)))))))))))))

By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: 

   a(a(a(a(b(x))))) -> b(a(b(a(b(a(a(b(a(a(a(a(x))))))))))))

----------------------------------------

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a(a(a(a(b(x))))) -> b(a(b(a(b(a(a(b(a(a(a(a(x))))))))))))

Q is empty.

----------------------------------------

(5) RFCMatchBoundsTRSProof (EQUIVALENT)
Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R.
The following rules were used to construct the certificate:

   a(a(a(a(b(x))))) -> b(a(b(a(b(a(a(b(a(a(a(a(x))))))))))))

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123

Node 100 is start node and node 101 is final node.

Those nodes are connected through the following edges:

* 100 to 102 labelled b_1(0)* 101 to 101 labelled #_1(0)* 102 to 103 labelled a_1(0)* 103 to 104 labelled b_1(0)* 104 to 105 labelled a_1(0)* 105 to 106 labelled b_1(0)* 106 to 107 labelled a_1(0)* 107 to 108 labelled a_1(0)* 108 to 109 labelled b_1(0)* 109 to 110 labelled a_1(0)* 109 to 113 labelled b_1(1)* 110 to 111 labelled a_1(0)* 110 to 113 labelled b_1(1)* 111 to 112 labelled a_1(0)* 111 to 113 labelled b_1(1)* 112 to 101 labelled a_1(0)* 112 to 113 labelled b_1(1)* 113 to 114 labelled a_1(1)* 114 to 115 labelled b_1(1)* 115 to 116 labelled a_1(1)* 116 to 117 labelled b_1(1)* 117 to 118 labelled a_1(1)* 118 to 119 labelled a_1(1)* 119 to 120 labelled b_1(1)* 120 to 121 labelled a_1(1)* 120 to 113 labelled b_1(1)* 121 to 122 labelled a_1(1)* 121 to 113 labelled b_1(1)* 122 to 123 labelled a_1(1)* 122 to 113 labelled b_1(1)* 123 to 101 labelled a_1(1)* 123 to 113 labelled b_1(1)


----------------------------------------

(6)
YES