Certification Problem
Input (TPDB TRS_Standard/Secret_07_TRS/5)
The rewrite relation of the following TRS is considered.
|
g(x,x,x) |
→ |
g(c,d,e) |
(1) |
|
g(x,y,x) |
→ |
g(c,d,e) |
(2) |
|
s(f(x,y)) |
→ |
f(y,f(s(s(x)),a)) |
(3) |
|
h(h(x,a),y) |
→ |
h(h(a,y),h(a,x)) |
(4) |
|
f(x,f(y,f(x,y))) |
→ |
f(a,f(x,f(y,b))) |
(5) |
|
f(h(a,y),g(x,b,a)) |
→ |
h(f(x,s(y)),s(b)) |
(6) |
|
h(f(x,s(y)),b) |
→ |
f(a,g(y,a,f(s(x),a))) |
(7) |
|
f(x,g(x,a,f(s(x),y))) |
→ |
f(h(x,b),g(a,b,y)) |
(8) |
|
s(y) |
→ |
b |
(9) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
g#(x,x,x) |
→ |
g#(c,d,e) |
(10) |
|
g#(x,y,x) |
→ |
g#(c,d,e) |
(11) |
|
s#(f(x,y)) |
→ |
f#(y,f(s(s(x)),a)) |
(12) |
|
s#(f(x,y)) |
→ |
f#(s(s(x)),a) |
(13) |
|
s#(f(x,y)) |
→ |
s#(s(x)) |
(14) |
|
s#(f(x,y)) |
→ |
s#(x) |
(15) |
|
h#(h(x,a),y) |
→ |
h#(h(a,y),h(a,x)) |
(16) |
|
h#(h(x,a),y) |
→ |
h#(a,y) |
(17) |
|
h#(h(x,a),y) |
→ |
h#(a,x) |
(18) |
|
f#(x,f(y,f(x,y))) |
→ |
f#(a,f(x,f(y,b))) |
(19) |
|
f#(x,f(y,f(x,y))) |
→ |
f#(x,f(y,b)) |
(20) |
|
f#(x,f(y,f(x,y))) |
→ |
f#(y,b) |
(21) |
|
f#(h(a,y),g(x,b,a)) |
→ |
h#(f(x,s(y)),s(b)) |
(22) |
|
f#(h(a,y),g(x,b,a)) |
→ |
f#(x,s(y)) |
(23) |
|
f#(h(a,y),g(x,b,a)) |
→ |
s#(y) |
(24) |
|
f#(h(a,y),g(x,b,a)) |
→ |
s#(b) |
(25) |
|
h#(f(x,s(y)),b) |
→ |
f#(a,g(y,a,f(s(x),a))) |
(26) |
|
h#(f(x,s(y)),b) |
→ |
g#(y,a,f(s(x),a)) |
(27) |
|
h#(f(x,s(y)),b) |
→ |
f#(s(x),a) |
(28) |
|
h#(f(x,s(y)),b) |
→ |
s#(x) |
(29) |
|
f#(x,g(x,a,f(s(x),y))) |
→ |
f#(h(x,b),g(a,b,y)) |
(30) |
|
f#(x,g(x,a,f(s(x),y))) |
→ |
h#(x,b) |
(31) |
|
f#(x,g(x,a,f(s(x),y))) |
→ |
g#(a,b,y) |
(32) |
1.1 Dependency Graph Processor
The dependency pairs are split into 4
components.
-
The
1st
component contains the
pair
|
h#(f(x,s(y)),b) |
→ |
f#(a,g(y,a,f(s(x),a))) |
(26) |
|
f#(x,g(x,a,f(s(x),y))) |
→ |
f#(h(x,b),g(a,b,y)) |
(30) |
|
f#(h(a,y),g(x,b,a)) |
→ |
h#(f(x,s(y)),s(b)) |
(22) |
|
f#(x,g(x,a,f(s(x),y))) |
→ |
h#(x,b) |
(31) |
1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [s(x1)] |
= |
1 · x1
|
| [f(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [a] |
= |
0 |
| [b] |
= |
0 |
| [h(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [g(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
| [c] |
= |
0 |
| [d] |
= |
0 |
| [e] |
= |
0 |
| [f#(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [h#(x1, x2)] |
= |
1 · x1 + 1 · x2
|
together with the usable
rules
|
s(f(x,y)) |
→ |
f(y,f(s(s(x)),a)) |
(3) |
|
s(y) |
→ |
b |
(9) |
|
f(x,f(y,f(x,y))) |
→ |
f(a,f(x,f(y,b))) |
(5) |
|
h(f(x,s(y)),b) |
→ |
f(a,g(y,a,f(s(x),a))) |
(7) |
|
f(x,g(x,a,f(s(x),y))) |
→ |
f(h(x,b),g(a,b,y)) |
(8) |
|
f(h(a,y),g(x,b,a)) |
→ |
h(f(x,s(y)),s(b)) |
(6) |
|
g(x,y,x) |
→ |
g(c,d,e) |
(2) |
|
h(h(x,a),y) |
→ |
h(h(a,y),h(a,x)) |
(4) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
1.1.1.1 Instantiation Processor
We instantiate the pair
to the following set of pairs
|
f#(a,g(a,a,f(s(a),x1))) |
→ |
f#(h(a,b),g(a,b,x1)) |
(33) |
1.1.1.1.1 Instantiation Processor
We instantiate the pair
to the following set of pairs
|
f#(h(a,b),g(x1,b,a)) |
→ |
h#(f(x1,s(b)),s(b)) |
(34) |
1.1.1.1.1.1 Instantiation Processor
We instantiate the pair
to the following set of pairs
|
f#(a,g(a,a,f(s(a),x1))) |
→ |
h#(a,b) |
(35) |
1.1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
2nd
component contains the
pair
|
h#(h(x,a),y) |
→ |
h#(h(a,y),h(a,x)) |
(16) |
1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [h(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [a] |
= |
0 |
| [h#(x1, x2)] |
= |
1 · x1 + 1 · x2
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
rule
could be deleted.
1.1.2.1 Reduction Pair Processor
Using the linear polynomial interpretation over the rationals with delta = 1/8
| [h#(x1, x2)] |
= |
0 + 1 · x1 + 1/2 · x2
|
| [h(x1, x2)] |
= |
0 + 1/4 · x1 + 1/2 · x2
|
| [a] |
= |
1 |
the
pair
|
h#(h(x,a),y) |
→ |
h#(h(a,y),h(a,x)) |
(16) |
could be deleted.
1.1.2.1.1 P is empty
There are no pairs anymore.
-
The
3rd
component contains the
pair
|
s#(f(x,y)) |
→ |
s#(x) |
(15) |
|
s#(f(x,y)) |
→ |
s#(s(x)) |
(14) |
1.1.3 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [s(x1)] |
= |
1 · x1
|
| [f(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [a] |
= |
0 |
| [b] |
= |
0 |
| [s#(x1)] |
= |
1 · x1
|
together with the usable
rules
|
s(f(x,y)) |
→ |
f(y,f(s(s(x)),a)) |
(3) |
|
s(y) |
→ |
b |
(9) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
1.1.3.1 Narrowing Processor
We consider all narrowings of the pair
below position
1
to get the following set of pairs
|
s#(f(f(x0,x1),y1)) |
→ |
s#(f(x1,f(s(s(x0)),a))) |
(39) |
|
s#(f(x0,y1)) |
→ |
s#(b) |
(40) |
1.1.3.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
4th
component contains the
pair
|
f#(x,f(y,f(x,y))) |
→ |
f#(a,f(x,f(y,b))) |
(19) |
1.1.4 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [f(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [a] |
= |
0 |
| [b] |
= |
0 |
| [f#(x1, x2)] |
= |
1 · x1 + 1 · x2
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
rule
could be deleted.
1.1.4.1 Instantiation Processor
We instantiate the pair
to the following set of pairs
|
f#(a,f(b,f(a,b))) |
→ |
f#(a,f(a,f(b,b))) |
(41) |
1.1.4.1.1 Instantiation Processor
We instantiate the pair
to the following set of pairs
There are no rules.
1.1.4.1.1.1 P is empty
There are no pairs anymore.