Certification Problem
Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-2-num-15)
The rewrite relation of the following TRS is considered.
a(x1) |
→ |
x1 |
(1) |
a(a(x1)) |
→ |
a(b(a(b(x1)))) |
(2) |
b(b(b(b(x1)))) |
→ |
a(x1) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(x1) |
→ |
x1 |
(1) |
a(a(x1)) |
→ |
b(a(b(a(x1)))) |
(4) |
b(b(b(b(x1)))) |
→ |
a(x1) |
(3) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{a(☐), b(☐)}
We obtain the transformed TRS
a(a(x1)) |
→ |
a(x1) |
(5) |
b(a(x1)) |
→ |
b(x1) |
(6) |
a(a(a(x1))) |
→ |
a(b(a(b(a(x1))))) |
(7) |
b(a(a(x1))) |
→ |
b(b(a(b(a(x1))))) |
(8) |
a(b(b(b(b(x1))))) |
→ |
a(a(x1)) |
(9) |
b(b(b(b(b(x1))))) |
→ |
b(a(x1)) |
(10) |
1.1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
aa(aa(x1)) |
→ |
aa(x1) |
(11) |
aa(ab(x1)) |
→ |
ab(x1) |
(12) |
ba(aa(x1)) |
→ |
ba(x1) |
(13) |
ba(ab(x1)) |
→ |
bb(x1) |
(14) |
aa(aa(aa(x1))) |
→ |
ab(ba(ab(ba(aa(x1))))) |
(15) |
aa(aa(ab(x1))) |
→ |
ab(ba(ab(ba(ab(x1))))) |
(16) |
ba(aa(aa(x1))) |
→ |
bb(ba(ab(ba(aa(x1))))) |
(17) |
ba(aa(ab(x1))) |
→ |
bb(ba(ab(ba(ab(x1))))) |
(18) |
ab(bb(bb(bb(ba(x1))))) |
→ |
aa(aa(x1)) |
(19) |
ab(bb(bb(bb(bb(x1))))) |
→ |
aa(ab(x1)) |
(20) |
bb(bb(bb(bb(ba(x1))))) |
→ |
ba(aa(x1)) |
(21) |
bb(bb(bb(bb(bb(x1))))) |
→ |
ba(ab(x1)) |
(22) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[aa(x1)] |
= |
1 · x1 + 2 |
[ab(x1)] |
= |
1 · x1
|
[ba(x1)] |
= |
1 · x1 + 1 |
[bb(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
aa(aa(x1)) |
→ |
aa(x1) |
(11) |
aa(ab(x1)) |
→ |
ab(x1) |
(12) |
ba(aa(x1)) |
→ |
ba(x1) |
(13) |
aa(aa(aa(x1))) |
→ |
ab(ba(ab(ba(aa(x1))))) |
(15) |
aa(aa(ab(x1))) |
→ |
ab(ba(ab(ba(ab(x1))))) |
(16) |
ab(bb(bb(bb(bb(x1))))) |
→ |
aa(ab(x1)) |
(20) |
bb(bb(bb(bb(ba(x1))))) |
→ |
ba(aa(x1)) |
(21) |
bb(bb(bb(bb(bb(x1))))) |
→ |
ba(ab(x1)) |
(22) |
1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
ba#(aa(aa(x1))) |
→ |
ba#(ab(ba(aa(x1)))) |
(23) |
ba#(aa(aa(x1))) |
→ |
ab#(ba(aa(x1))) |
(24) |
ba#(aa(aa(x1))) |
→ |
ba#(aa(x1)) |
(25) |
ba#(aa(ab(x1))) |
→ |
ba#(ab(ba(ab(x1)))) |
(26) |
ba#(aa(ab(x1))) |
→ |
ab#(ba(ab(x1))) |
(27) |
ba#(aa(ab(x1))) |
→ |
ba#(ab(x1)) |
(28) |
1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.