Platform claims not to support CPU time, falling back to wall time.
Cannot find out cpu time on external processes, falling back to wall time!
proof of /home/fs5/ayamada/tpdb/relative/Relative_05/rt1-5.trs
# AProVE Commit ID: 2b684c7cda508b1711f707cb42f127e64fe1df88 ffrohn 20140415 dirty
Termination of the given RelTRS could be proven:
(0) RelTRS
(1) RelTRSRRRProof [EQUIVALENT, 105 ms]
(2) RelTRS
(3) RelTRSRRRProof [EQUIVALENT, 118 ms]
(4) RelTRS
(5) RIsEmptyProof [EQUIVALENT, 3 ms]
(6) YES
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(0)
Obligation:
Relative term rewrite system:
The relative TRS consists of the following R rules:
s(a(x)) -> s(b(x))
b(b(x)) -> a(x)
The relative TRS consists of the following S rules:
f(s(x), y) -> f(x, s(y))
s(a(x)) -> a(s(x))
s(b(x)) -> b(s(x))
a(s(x)) -> s(a(x))
b(s(x)) -> s(b(x))
----------------------------------------
(1) RelTRSRRRProof (EQUIVALENT)
We used the following monotonic ordering for rule removal:
Polynomial interpretation [POLO]:
POL(a(x_1)) = 1 + x_1
POL(b(x_1)) = 1 + x_1
POL(f(x_1, x_2)) = x_1 + x_2
POL(s(x_1)) = x_1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:
b(b(x)) -> a(x)
Rules from S:
none
----------------------------------------
(2)
Obligation:
Relative term rewrite system:
The relative TRS consists of the following R rules:
s(a(x)) -> s(b(x))
The relative TRS consists of the following S rules:
f(s(x), y) -> f(x, s(y))
s(a(x)) -> a(s(x))
s(b(x)) -> b(s(x))
a(s(x)) -> s(a(x))
b(s(x)) -> s(b(x))
----------------------------------------
(3) RelTRSRRRProof (EQUIVALENT)
We used the following monotonic ordering for rule removal:
Polynomial interpretation [POLO]:
POL(a(x_1)) = 1 + x_1
POL(b(x_1)) = x_1
POL(f(x_1, x_2)) = x_1 + x_2
POL(s(x_1)) = x_1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:
s(a(x)) -> s(b(x))
Rules from S:
none
----------------------------------------
(4)
Obligation:
Relative term rewrite system:
R is empty.
The relative TRS consists of the following S rules:
f(s(x), y) -> f(x, s(y))
s(a(x)) -> a(s(x))
s(b(x)) -> b(s(x))
a(s(x)) -> s(a(x))
b(s(x)) -> s(b(x))
----------------------------------------
(5) RIsEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
----------------------------------------
(6)
YES