Certification Problem
Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-223)
The rewrite relation of the following TRS is considered.
a(x1) |
→ |
x1 |
(1) |
a(b(x1)) |
→ |
b(b(c(a(c(x1))))) |
(2) |
b(x1) |
→ |
x1 |
(3) |
c(c(x1)) |
→ |
a(x1) |
(4) |
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) |
b(a(x1)) |
→ |
c(a(c(b(b(x1))))) |
(5) |
b(x1) |
→ |
x1 |
(3) |
c(c(x1)) |
→ |
a(x1) |
(4) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
b#(a(x1)) |
→ |
c#(a(c(b(b(x1))))) |
(6) |
b#(a(x1)) |
→ |
a#(c(b(b(x1)))) |
(7) |
b#(a(x1)) |
→ |
c#(b(b(x1))) |
(8) |
b#(a(x1)) |
→ |
b#(b(x1)) |
(9) |
b#(a(x1)) |
→ |
b#(x1) |
(10) |
c#(c(x1)) |
→ |
a#(x1) |
(11) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
1st
component contains the
pair
b#(a(x1)) |
→ |
b#(x1) |
(10) |
b#(a(x1)) |
→ |
b#(b(x1)) |
(9) |
1.1.1.1 Semantic Labeling Processor
The following interpretations form a
model
of the rules.
As carrier we take the set
{0,1}.
Symbols are labeled by the interpretation of their arguments using the interpretations
(modulo 2):
[a(x1)] |
= |
1x1
|
[b(x1)] |
= |
1x1
|
[c(x1)] |
= |
1 + 1x1
|
[b#(x1)] |
= |
0 |
We obtain the set of labeled pairs
b#0(a0(x1)) |
→ |
b#0(x1) |
(12) |
b#0(a0(x1)) |
→ |
b#0(b0(x1)) |
(13) |
b#1(a1(x1)) |
→ |
b#1(b1(x1)) |
(14) |
b#1(a1(x1)) |
→ |
b#1(x1) |
(15) |
and the set of labeled rules:
a0(x1) |
→ |
x1 |
(16) |
a1(x1) |
→ |
x1 |
(17) |
b0(a0(x1)) |
→ |
c1(a1(c0(b0(b0(x1))))) |
(18) |
b1(a1(x1)) |
→ |
c0(a0(c1(b1(b1(x1))))) |
(19) |
b0(x1) |
→ |
x1 |
(20) |
b1(x1) |
→ |
x1 |
(21) |
c1(c0(x1)) |
→ |
a0(x1) |
(22) |
c0(c1(x1)) |
→ |
a1(x1) |
(23) |
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.