Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
      ↳19 Instantiation (⇔)
        ↳20 QDP
      ↳21 NonInfProof (⇔)
        ↳22 QDP
        ↳23 DependencyGraphProof (⇔)
        ↳24 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
minus(x0, x1)
help(true, x0, x1)
help(false, x0, x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LT(s(x), s(y)) → LT(x, y)
MINUS(x, y) → HELP(lt(y, x), x, y)
MINUS(x, y) → LT(y, x)
HELP(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
minus(x0, x1)
help(true, x0, x1)
help(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
minus(x0, x1)
help(true, x0, x1)
help(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LT(s(x), s(y)) → LT(x, y)

R is empty.
The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
minus(x0, x1)
help(true, x0, x1)
help(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
minus(x0, x1)
help(true, x0, x1)
help(false, x0, x1)

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LT(s(x), s(y)) → LT(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • LT(s(x), s(y)) → LT(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(x, y) → HELP(lt(y, x), x, y)
HELP(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
minus(x, y) → help(lt(y, x), x, y)
help(true, x, y) → s(minus(x, s(y)))
help(false, x, y) → 0

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
minus(x0, x1)
help(true, x0, x1)
help(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(x, y) → HELP(lt(y, x), x, y)
HELP(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
minus(x0, x1)
help(true, x0, x1)
help(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, x1)
help(true, x0, x1)
help(false, x0, x1)

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(x, y) → HELP(lt(y, x), x, y)
HELP(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(19) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule MINUS(x, y) → HELP(lt(y, x), x, y) we obtained the following new rules [LPAR04]:

MINUS(z0, s(z1)) → HELP(lt(s(z1), z0), z0, s(z1))

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

HELP(true, x, y) → MINUS(x, s(y))
MINUS(z0, s(z1)) → HELP(lt(s(z1), z0), z0, s(z1))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(21) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair MINUS(x, y) → HELP(lt(y, x), x, y) the following chains were created:
  • We consider the chain HELP(true, x, y) → MINUS(x, s(y)), MINUS(x, y) → HELP(lt(y, x), x, y) which results in the following constraint:

    (1)    (MINUS(x2, s(x3))=MINUS(x4, x5) ⇒ MINUS(x4, x5)≥HELP(lt(x5, x4), x4, x5))



    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:

    (2)    (MINUS(x2, s(x3))≥HELP(lt(s(x3), x2), x2, s(x3)))







For Pair HELP(true, x, y) → MINUS(x, s(y)) the following chains were created:
  • We consider the chain MINUS(x, y) → HELP(lt(y, x), x, y), HELP(true, x, y) → MINUS(x, s(y)) which results in the following constraint:

    (3)    (HELP(lt(x7, x6), x6, x7)=HELP(true, x8, x9) ⇒ HELP(true, x8, x9)≥MINUS(x8, s(x9)))



    We simplified constraint (3) using rules (I), (II), (III) which results in the following new constraint:

    (4)    (lt(x7, x6)=true ⇒ HELP(true, x6, x7)≥MINUS(x6, s(x7)))



    We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on lt(x7, x6)=true which results in the following new constraints:

    (5)    (true=true ⇒ HELP(true, s(x12), 0)≥MINUS(s(x12), s(0)))


    (6)    (lt(x15, x14)=true∧(lt(x15, x14)=true ⇒ HELP(true, x14, x15)≥MINUS(x14, s(x15))) ⇒ HELP(true, s(x14), s(x15))≥MINUS(s(x14), s(s(x15))))



    We simplified constraint (5) using rules (I), (II) which results in the following new constraint:

    (7)    (HELP(true, s(x12), 0)≥MINUS(s(x12), s(0)))



    We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (lt(x15, x14)=true ⇒ HELP(true, x14, x15)≥MINUS(x14, s(x15))) with σ = [ ] which results in the following new constraint:

    (8)    (HELP(true, x14, x15)≥MINUS(x14, s(x15)) ⇒ HELP(true, s(x14), s(x15))≥MINUS(s(x14), s(s(x15))))







To summarize, we get the following constraints P≥ for the following pairs.
  • MINUS(x, y) → HELP(lt(y, x), x, y)
    • (MINUS(x2, s(x3))≥HELP(lt(s(x3), x2), x2, s(x3)))

  • HELP(true, x, y) → MINUS(x, s(y))
    • (HELP(true, s(x12), 0)≥MINUS(s(x12), s(0)))
    • (HELP(true, x14, x15)≥MINUS(x14, s(x15)) ⇒ HELP(true, s(x14), s(x15))≥MINUS(s(x14), s(s(x15))))




The constraints for P> respective Pbound are constructed from P≥ where we just replace every occurence of "t ≥ s" in P≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(HELP(x1, x2, x3)) = -1 - x1 + x2 - x3   
POL(MINUS(x1, x2)) = -1 + x1 - x2   
POL(c) = -2   
POL(false) = 0   
POL(lt(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following pairs are in P>:

HELP(true, x, y) → MINUS(x, s(y))
The following pairs are in Pbound:

HELP(true, x, y) → MINUS(x, s(y))
The following rules are usable:

true → lt(0, s(x))
false → lt(x, 0)
lt(x, y) → lt(s(x), s(y))

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(x, y) → HELP(lt(y, x), x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(24) TRUE