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

(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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

(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)
TIMES(s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → TIMES(x, y)
PLUS(s(x), y) → PLUS(x, y)
FAC(x) → LOOP(x, s(0), s(0))
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
LOOP(x, c, y) → LT(x, c)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
IF(false, x, c, y) → TIMES(y, s(c))

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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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:

PLUS(s(x), y) → PLUS(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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

(11) Obligation:

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

PLUS(s(x), y) → PLUS(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:

• PLUS(s(x), y) → PLUS(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:

TIMES(s(x), y) → TIMES(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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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:

TIMES(s(x), y) → TIMES(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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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].

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

(18) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

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

(19) 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:

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

(20) TRUE

(21) 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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(22) 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.

(23) 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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(24) 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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

(25) 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.

(26) 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

(27) TRUE

(28) Obligation:

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

LOOP(x, c, y) → IF(lt(x, c), x, c, y)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))

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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(29) 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.

(30) Obligation:

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

LOOP(x, c, y) → IF(lt(x, c), x, c, y)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))

The TRS R consists of the following rules:

times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

(31) 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].

fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

(32) Obligation:

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

LOOP(x, c, y) → IF(lt(x, c), x, c, y)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))

The TRS R consists of the following rules:

times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

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

(33) Instantiation (EQUIVALENT transformation)

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

LOOP(z0, s(z1), y_0) → IF(lt(z0, s(z1)), z0, s(z1), y_0)

(34) Obligation:

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

IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
LOOP(z0, s(z1), y_0) → IF(lt(z0, s(z1)), z0, s(z1), y_0)

The TRS R consists of the following rules:

times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

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

(35) 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 LOOP(x, c, y) → IF(lt(x, c), x, c, y) the following chains were created:

• We consider the chain IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c))), LOOP(x, c, y) → IF(lt(x, c), x, c, y) which results in the following constraint:
  

  (1)    (LOOP(x3, s(x4), times(x5, s(x4)))=LOOP(x6, x7, x8) ⇒ LOOP(x6, x7, x8)≥IF(lt(x6, x7), x6, x7, x8))

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

  (2)    (LOOP(x3, s(x4), x8)≥IF(lt(x3, s(x4)), x3, s(x4), x8))

  
  
  

For Pair IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c))) the following chains were created:

• We consider the chain LOOP(x, c, y) → IF(lt(x, c), x, c, y), IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c))) which results in the following constraint:
  

  (3)    (IF(lt(x9, x10), x9, x10, x11)=IF(false, x12, x13, x14) ⇒ IF(false, x12, x13, x14)≥LOOP(x12, s(x13), times(x14, s(x13))))

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

  (4)    (lt(x9, x10)=false ⇒ IF(false, x9, x10, x11)≥LOOP(x9, s(x10), times(x11, s(x10))))

  
  
  We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on lt(x9, x10)=false which results in the following new constraints:
  

  (5)    (false=false ⇒ IF(false, x20, 0, x11)≥LOOP(x20, s(0), times(x11, s(0))))

  
  

  (6)    (lt(x22, x21)=false∧(∀x23:lt(x22, x21)=false ⇒ IF(false, x22, x21, x23)≥LOOP(x22, s(x21), times(x23, s(x21)))) ⇒ IF(false, s(x22), s(x21), x11)≥LOOP(s(x22), s(s(x21)), times(x11, s(s(x21)))))

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

  (7)    (IF(false, x20, 0, x11)≥LOOP(x20, s(0), times(x11, s(0))))

  
  
  We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (∀x23:lt(x22, x21)=false ⇒ IF(false, x22, x21, x23)≥LOOP(x22, s(x21), times(x23, s(x21)))) with σ = [x23 / x11] which results in the following new constraint:
  

  (8)    (IF(false, x22, x21, x11)≥LOOP(x22, s(x21), times(x11, s(x21))) ⇒ IF(false, s(x22), s(x21), x11)≥LOOP(s(x22), s(s(x21)), times(x11, s(s(x21)))))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• LOOP(x, c, y) → IF(lt(x, c), x, c, y)
  
  • (LOOP(x3, s(x4), x8)≥IF(lt(x3, s(x4)), x3, s(x4), x8))
    
  
  
• IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
  
  • (IF(false, x20, 0, x11)≥LOOP(x20, s(0), times(x11, s(0))))
    
  • (IF(false, x22, x21, x11)≥LOOP(x22, s(x21), times(x11, s(x21))) ⇒ IF(false, s(x22), s(x21), x11)≥LOOP(s(x22), s(s(x21)), times(x11, s(s(x21)))))
    
  
  

The constraints for P_> respective P_bound 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 P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(IF(x₁, x₂, x₃, x₄)) = 1 + x₂ - x₃   
POL(LOOP(x₁, x₂, x₃)) = 1 + x₁ - x₂   
POL(c) = -1   
POL(false) = 1   
POL(lt(x₁, x₂)) = 0   
POL(plus(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(times(x₁, x₂)) = 1 + x₂   
POL(true) = 0   

The following pairs are in P_>:

IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))

The following pairs are in P_bound:

IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))

There are no usable rules

(36) Obligation:

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

LOOP(x, c, y) → IF(lt(x, c), x, c, y)

The TRS R consists of the following rules:

times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

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

(37) DependencyGraphProof (EQUIVALENT transformation)

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

(38) TRUE