Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-254)
The rewrite relation of the following TRS is considered.
a(a(b(b(x1)))) |
→ |
a(a(a(b(x1)))) |
(1) |
b(b(a(b(x1)))) |
→ |
a(a(b(a(x1)))) |
(2) |
b(a(a(a(x1)))) |
→ |
b(a(b(b(x1)))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Closure Under Flat Contexts
Using the flat contexts
{a(☐), b(☐)}
We obtain the transformed TRS
a(a(b(b(x1)))) |
→ |
a(a(a(b(x1)))) |
(1) |
a(b(b(a(b(x1))))) |
→ |
a(a(a(b(a(x1))))) |
(4) |
b(b(b(a(b(x1))))) |
→ |
b(a(a(b(a(x1))))) |
(5) |
b(a(a(a(x1)))) |
→ |
b(a(b(b(x1)))) |
(3) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
aa(ab(bb(ba(x1)))) |
→ |
aa(aa(ab(ba(x1)))) |
(6) |
aa(ab(bb(bb(x1)))) |
→ |
aa(aa(ab(bb(x1)))) |
(7) |
ab(bb(ba(ab(ba(x1))))) |
→ |
aa(aa(ab(ba(aa(x1))))) |
(8) |
ab(bb(ba(ab(bb(x1))))) |
→ |
aa(aa(ab(ba(ab(x1))))) |
(9) |
bb(bb(ba(ab(ba(x1))))) |
→ |
ba(aa(ab(ba(aa(x1))))) |
(10) |
bb(bb(ba(ab(bb(x1))))) |
→ |
ba(aa(ab(ba(ab(x1))))) |
(11) |
ba(aa(aa(aa(x1)))) |
→ |
ba(ab(bb(ba(x1)))) |
(12) |
ba(aa(aa(ab(x1)))) |
→ |
ba(ab(bb(bb(x1)))) |
(13) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
aa#(ab(bb(ba(x1)))) |
→ |
ab#(ba(x1)) |
(14) |
aa#(ab(bb(ba(x1)))) |
→ |
aa#(ab(ba(x1))) |
(15) |
aa#(ab(bb(ba(x1)))) |
→ |
aa#(aa(ab(ba(x1)))) |
(16) |
aa#(ab(bb(bb(x1)))) |
→ |
ab#(bb(x1)) |
(17) |
aa#(ab(bb(bb(x1)))) |
→ |
aa#(ab(bb(x1))) |
(18) |
aa#(ab(bb(bb(x1)))) |
→ |
aa#(aa(ab(bb(x1)))) |
(19) |
ab#(bb(ba(ab(ba(x1))))) |
→ |
aa#(x1) |
(20) |
ab#(bb(ba(ab(ba(x1))))) |
→ |
ba#(aa(x1)) |
(21) |
ab#(bb(ba(ab(ba(x1))))) |
→ |
ab#(ba(aa(x1))) |
(22) |
ab#(bb(ba(ab(ba(x1))))) |
→ |
aa#(ab(ba(aa(x1)))) |
(23) |
ab#(bb(ba(ab(ba(x1))))) |
→ |
aa#(aa(ab(ba(aa(x1))))) |
(24) |
ab#(bb(ba(ab(bb(x1))))) |
→ |
ab#(x1) |
(25) |
ab#(bb(ba(ab(bb(x1))))) |
→ |
ba#(ab(x1)) |
(26) |
ab#(bb(ba(ab(bb(x1))))) |
→ |
ab#(ba(ab(x1))) |
(27) |
ab#(bb(ba(ab(bb(x1))))) |
→ |
aa#(ab(ba(ab(x1)))) |
(28) |
ab#(bb(ba(ab(bb(x1))))) |
→ |
aa#(aa(ab(ba(ab(x1))))) |
(29) |
bb#(bb(ba(ab(ba(x1))))) |
→ |
aa#(x1) |
(30) |
bb#(bb(ba(ab(ba(x1))))) |
→ |
ba#(aa(x1)) |
(31) |
bb#(bb(ba(ab(ba(x1))))) |
→ |
ab#(ba(aa(x1))) |
(32) |
bb#(bb(ba(ab(ba(x1))))) |
→ |
aa#(ab(ba(aa(x1)))) |
(33) |
bb#(bb(ba(ab(ba(x1))))) |
→ |
ba#(aa(ab(ba(aa(x1))))) |
(34) |
bb#(bb(ba(ab(bb(x1))))) |
→ |
ab#(x1) |
(35) |
bb#(bb(ba(ab(bb(x1))))) |
→ |
ba#(ab(x1)) |
(36) |
bb#(bb(ba(ab(bb(x1))))) |
→ |
ab#(ba(ab(x1))) |
(37) |
bb#(bb(ba(ab(bb(x1))))) |
→ |
aa#(ab(ba(ab(x1)))) |
(38) |
bb#(bb(ba(ab(bb(x1))))) |
→ |
ba#(aa(ab(ba(ab(x1))))) |
(39) |
ba#(aa(aa(aa(x1)))) |
→ |
ba#(x1) |
(40) |
ba#(aa(aa(aa(x1)))) |
→ |
bb#(ba(x1)) |
(41) |
ba#(aa(aa(aa(x1)))) |
→ |
ab#(bb(ba(x1))) |
(42) |
ba#(aa(aa(aa(x1)))) |
→ |
ba#(ab(bb(ba(x1)))) |
(43) |
ba#(aa(aa(ab(x1)))) |
→ |
bb#(x1) |
(44) |
ba#(aa(aa(ab(x1)))) |
→ |
bb#(bb(x1)) |
(45) |
ba#(aa(aa(ab(x1)))) |
→ |
ab#(bb(bb(x1))) |
(46) |
ba#(aa(aa(ab(x1)))) |
→ |
ba#(ab(bb(bb(x1)))) |
(47) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.