The rewrite relation of the following TRS is considered.
a(a(a(b(b(b(x1)))))) | → | b(b(b(b(a(a(a(a(x1)))))))) | (1) |
b(b(b(a(a(a(x1)))))) | → | a(a(a(a(b(b(b(b(x1)))))))) | (2) |
{b(☐), a(☐)}
We obtain the transformed TRSb(b(b(b(a(a(a(x1))))))) | → | b(a(a(a(a(b(b(b(b(x1))))))))) | (3) |
a(b(b(b(a(a(a(x1))))))) | → | a(a(a(a(a(b(b(b(b(x1))))))))) | (4) |
Root-labeling is applied.
We obtain the labeled TRSbb(bb(bb(ba(aa(aa(ab(x1))))))) | → | ba(aa(aa(aa(ab(bb(bb(bb(bb(x1))))))))) | (5) |
bb(bb(bb(ba(aa(aa(aa(x1))))))) | → | ba(aa(aa(aa(ab(bb(bb(bb(ba(x1))))))))) | (6) |
ab(bb(bb(ba(aa(aa(ab(x1))))))) | → | aa(aa(aa(aa(ab(bb(bb(bb(bb(x1))))))))) | (7) |
ab(bb(bb(ba(aa(aa(aa(x1))))))) | → | aa(aa(aa(aa(ab(bb(bb(bb(ba(x1))))))))) | (8) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | ab#(bb(bb(bb(bb(x1))))) | (9) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(bb(bb(x1)))) | (10) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(bb(x1))) | (11) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(x1)) | (12) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(x1) | (13) |
bb#(bb(bb(ba(aa(aa(aa(x1))))))) | → | ab#(bb(bb(bb(ba(x1))))) | (14) |
bb#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(bb(bb(ba(x1)))) | (15) |
bb#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(bb(ba(x1))) | (16) |
bb#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(ba(x1)) | (17) |
ab#(bb(bb(ba(aa(aa(ab(x1))))))) | → | ab#(bb(bb(bb(bb(x1))))) | (18) |
ab#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(bb(bb(x1)))) | (19) |
ab#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(bb(x1))) | (20) |
ab#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(x1)) | (21) |
ab#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(x1) | (22) |
ab#(bb(bb(ba(aa(aa(aa(x1))))))) | → | ab#(bb(bb(bb(ba(x1))))) | (23) |
ab#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(bb(bb(ba(x1)))) | (24) |
ab#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(bb(ba(x1))) | (25) |
ab#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(ba(x1)) | (26) |
The dependency pairs are split into 1 component.
ab#(bb(bb(ba(aa(aa(aa(x1))))))) | → | ab#(bb(bb(bb(ba(x1))))) | (23) |
ab#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(bb(bb(ba(x1)))) | (24) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | ab#(bb(bb(bb(bb(x1))))) | (9) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(bb(bb(x1)))) | (10) |
bb#(bb(bb(ba(aa(aa(aa(x1))))))) | → | ab#(bb(bb(bb(ba(x1))))) | (14) |
bb#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(bb(bb(ba(x1)))) | (15) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(bb(x1))) | (11) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(x1)) | (12) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(x1) | (13) |
[ab#(x1)] | = |
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[bb(x1)] | = |
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[ba(x1)] | = |
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[aa(x1)] | = |
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[bb#(x1)] | = |
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[ab(x1)] | = |
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ab#(bb(bb(ba(aa(aa(aa(x1))))))) | → | ab#(bb(bb(bb(ba(x1))))) | (23) |
bb#(bb(bb(ba(aa(aa(aa(x1))))))) | → | ab#(bb(bb(bb(ba(x1))))) | (14) |
[ab#(x1)] | = |
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[bb(x1)] | = |
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[ba(x1)] | = |
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[aa(x1)] | = |
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[bb#(x1)] | = |
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[ab(x1)] | = |
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ab#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(bb(bb(ba(x1)))) | (24) |
The dependency pairs are split into 1 component.
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(bb(bb(x1)))) | (10) |
bb#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(bb(bb(ba(x1)))) | (15) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(bb(x1))) | (11) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(x1)) | (12) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(x1) | (13) |
[bb#(x1)] | = |
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[bb(x1)] | = |
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[ba(x1)] | = |
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[aa(x1)] | = |
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[ab(x1)] | = |
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bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(bb(bb(x1)))) | (10) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(bb(x1))) | (11) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(bb(x1)) | (12) |
bb#(bb(bb(ba(aa(aa(ab(x1))))))) | → | bb#(x1) | (13) |
20
Hence, it suffices to show innermost termination in the following.We restrict the rewrite rules to the following usable rules of the DP problem.
There are no rules.
We restrict the innermost strategy to the following left hand sides.
bb(bb(bb(ba(aa(aa(ab(x0))))))) |
bb(bb(bb(ba(aa(aa(aa(x0))))))) |
[bb#(x1)] | = | x1 |
[bb(x1)] | = | -2 + 2 · x1 |
[ba(x1)] | = | x1 |
[aa(x1)] | = | 2 + 2 · x1 |
bb#(bb(bb(ba(aa(aa(aa(x1))))))) | → | bb#(bb(bb(ba(x1)))) | (15) |
There are no pairs anymore.