Certification Problem
Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.26)
The rewrite relation of the following TRS is considered.
p(0) |
→ |
0 |
(1) |
p(s(x)) |
→ |
x |
(2) |
le(0,y) |
→ |
true |
(3) |
le(s(x),0) |
→ |
false |
(4) |
le(s(x),s(y)) |
→ |
le(x,y) |
(5) |
minus(x,y) |
→ |
if(le(x,y),x,y) |
(6) |
if(true,x,y) |
→ |
0 |
(7) |
if(false,x,y) |
→ |
s(minus(p(x),y)) |
(8) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
le#(s(x),s(y)) |
→ |
le#(x,y) |
(9) |
minus#(x,y) |
→ |
le#(x,y) |
(10) |
minus#(x,y) |
→ |
if#(le(x,y),x,y) |
(11) |
if#(false,x,y) |
→ |
p#(x) |
(12) |
if#(false,x,y) |
→ |
minus#(p(x),y) |
(13) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
if#(false,x,y) |
→ |
minus#(p(x),y) |
(13) |
minus#(x,y) |
→ |
if#(le(x,y),x,y) |
(11) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the arctic semiring over the integers
[true] |
= |
0 |
[p(x1)] |
= |
-4 · x1 + 1 |
[le(x1, x2)] |
= |
-7 · x1 +
-∞ · x2 + 0 |
[false] |
= |
1 |
[minus#(x1, x2)] |
= |
-3 · x1 +
-∞ · x2 + 1 |
[0] |
= |
1 |
[if#(x1, x2, x3)] |
= |
0 · x1 + -7 · x2 +
-∞ · x3 + 0 |
[s(x1)] |
= |
6 · x1 + 11 |
together with the usable
rules
p(0) |
→ |
0 |
(1) |
p(s(x)) |
→ |
x |
(2) |
le(0,y) |
→ |
true |
(3) |
le(s(x),0) |
→ |
false |
(4) |
le(s(x),s(y)) |
→ |
le(x,y) |
(5) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pair
minus#(x,y) |
→ |
if#(le(x,y),x,y) |
(11) |
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
2nd
component contains the
pair
le#(s(x),s(y)) |
→ |
le#(x,y) |
(9) |
1.1.2 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
le#(s(x),s(y)) |
→ |
le#(x,y) |
(9) |
|
2 |
> |
2 |
1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.