Certification Problem

Input (TPDB TRS_Standard/AProVE_07/otto03)

The rewrite relation of the following TRS is considered.

lt(0,s(x))	→	true	(1)
lt(x,0)	→	false	(2)
lt(s(x),s(y))	→	lt(x,y)	(3)
fac(x)	→	help(x,0)	(4)
help(x,c)	→	if(lt(c,x),x,c)	(5)
if(true,x,c)	→	times(s(c),help(x,s(c)))	(6)
if(false,x,c)	→	s(0)	(7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

Hence, it suffices to show innermost termination in the following.

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

lt^#(s(x),s(y))	→	lt^#(x,y)	(8)
fac^#(x)	→	help^#(x,0)	(9)
help^#(x,c)	→	if^#(lt(c,x),x,c)	(10)
help^#(x,c)	→	lt^#(c,x)	(11)
if^#(true,x,c)	→	help^#(x,s(c))	(12)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

help^#(x,c)	→	if^#(lt(c,x),x,c)	(10)
if^#(true,x,c)	→	help^#(x,s(c))	(12)

1.1.1.1 Usable Rules Processor

We restrict the rewrite rules to the following usable rules of the DP problem.

lt(0,s(x))	→	true	(1)
lt(x,0)	→	false	(2)
lt(s(x),s(y))	→	lt(x,y)	(3)

1.1.1.1.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

lt(0,s(x0))

lt(x0,0)

lt(s(x0),s(x1))

1.1.1.1.1.1 Reduction Pair Processor

We apply the generic reduction pair processor using the linear polynomial interpretation over the naturals

[c]	=	-1
[if^#(x₁, x₂, x₃)]	=	-1 · x₁ + 1 · x₂ + -1 · x₃
[help^#(x₁, x₂)]	=	1 · x₁ + -1 · x₂ + 1
[lt(x₁, x₂)]	=	0
[true]	=	0
[s(x₁)]	=	1 · x₁ + 1
[0]	=	0
[false]	=	0

The pair(s)

help^#(x,c)

→

if^#(lt(c,x),x,c)

(10)

are strictly oriented and the pair(s)

if^#(true,x,c)

→

help^#(x,s(c))

(12)

are bounded w.r.t. the constant c.

The following constraints are generated for the pairs.

if^#(true,s(x12),0)≥c
(if^#(true,x14,x15)≥c⟹if^#(true,s(x14),s(x15))≥c)
help^#(x2,s(x3)) > if^#(lt(s(x3),x2),x2,s(x3))
if^#(true,s(x12),0)≥help^#(s(x12),s(0))
(if^#(true,x14,x15)≥help^#(x14,s(x15))⟹if^#(true,s(x14),s(x15))≥help^#(s(x14),s(s(x15))))

The details are shown below:

For the chain we build the initial constraint
(if^#(lt(x7,x6),x6,x7)→^*if^#(true,x8,x9)⟹if^#(true,x8,x9)≥c)

which is simplified as follows.
- Applying Rule "Same Constructor" results in
  (lt(x7,x6)→^*true⟹x6→^*x8⟹x7→^*x9⟹if^#(true,x8,x9)≥c)
- Applying Rule "Variable in Equation" allows to substitute x8 by x6 which results in
  (lt(x7,x6)→^*true⟹x7→^*x9⟹if^#(true,x6,x9)≥c)
- Applying Rule "Variable in Equation" allows to substitute x9 by x7 which results in
  (lt(x7,x6)→^*true⟹if^#(true,x6,x7)≥c)
- Applying Rule "Induction" on lt(x7,x6)→^*true results in the following 3 new constraints.
  1. (true→^*true⟹if^#(true,s(x12),0)≥c)
    - Applying Rule "Same Constructor" results in
      if^#(true,s(x12),0)≥c
    - This constraint is kept as final constraint.
  2. (false→^*true⟹if^#(true,0,x13)≥c)
    - Applying Rule "Different Constructors" allows to drop this constraint.
  3. (lt(x15,x14)→^*true⟹(lt(x15,x14)→^*true⟹if^#(true,x14,x15)≥c)⟹if^#(true,s(x14),s(x15))≥c)
    - Applying Rule "Simplify Conditions" results in
      (if^#(true,x14,x15)≥c⟹if^#(true,s(x14),s(x15))≥c)
    - This constraint is kept as final constraint.
For the chain we build the initial constraint
(help^#(x2,s(x3))→^*help^#(x4,x5)⟹help^#(x4,x5) > if^#(lt(x5,x4),x4,x5))

which is simplified as follows.
- Applying Rule "Same Constructor" results in
  (x2→^*x4⟹s(x3)→^*x5⟹help^#(x4,x5) > if^#(lt(x5,x4),x4,x5))
- Applying Rule "Variable in Equation" allows to substitute x4 by x2 which results in
  (s(x3)→^*x5⟹help^#(x2,x5) > if^#(lt(x5,x2),x2,x5))
- Applying Rule "Variable in Equation" allows to substitute x5 by s(x3) which results in
  help^#(x2,s(x3)) > if^#(lt(s(x3),x2),x2,s(x3))
- This constraint is kept as final constraint.
For the chain we build the initial constraint
(if^#(lt(x7,x6),x6,x7)→^*if^#(true,x8,x9)⟹if^#(true,x8,x9)≥help^#(x8,s(x9)))

which is simplified as follows.
- Applying Rule "Same Constructor" results in
  (lt(x7,x6)→^*true⟹x6→^*x8⟹x7→^*x9⟹if^#(true,x8,x9)≥help^#(x8,s(x9)))
- Applying Rule "Variable in Equation" allows to substitute x8 by x6 which results in
  (lt(x7,x6)→^*true⟹x7→^*x9⟹if^#(true,x6,x9)≥help^#(x6,s(x9)))
- Applying Rule "Variable in Equation" allows to substitute x9 by x7 which results in
  (lt(x7,x6)→^*true⟹if^#(true,x6,x7)≥help^#(x6,s(x7)))
- Applying Rule "Induction" on lt(x7,x6)→^*true results in the following 3 new constraints.
  1. (true→^*true⟹if^#(true,s(x12),0)≥help^#(s(x12),s(0)))
    - Applying Rule "Same Constructor" results in
      if^#(true,s(x12),0)≥help^#(s(x12),s(0))
    - This constraint is kept as final constraint.
  2. (false→^*true⟹if^#(true,0,x13)≥help^#(0,s(x13)))
    - Applying Rule "Different Constructors" allows to drop this constraint.
  3. (lt(x15,x14)→^*true⟹(lt(x15,x14)→^*true⟹if^#(true,x14,x15)≥help^#(x14,s(x15)))⟹if^#(true,s(x14),s(x15))≥help^#(s(x14),s(s(x15))))
    - Applying Rule "Simplify Conditions" results in
      (if^#(true,x14,x15)≥help^#(x14,s(x15))⟹if^#(true,s(x14),s(x15))≥help^#(s(x14),s(s(x15))))
    - This constraint is kept as final constraint.

We get two subproofs, in the first the strict pairs are deleted, in the second the bounded pairs are deleted.

1.1.1.1.1.1.1 Usable Rules Processor

We restrict the rewrite rules to the following usable rules of the DP problem.

There are no rules.

1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

There are no lhss.

1.1.1.1.1.1.1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

if^#(true,x,c)	→	help^#(x,s(c))	(12)

2	≥	1

As there is no critical graph in the transitive closure, there are no infinite chains.

1.1.1.1.1.1.2 Size-Change Termination