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:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, 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:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(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:

MINUS(x, y) → COND(equal(min(x, y), y), x, y)
MINUS(x, y) → EQUAL(min(x, y), y)
MINUS(x, y) → MIN(x, y)
COND(true, x, y) → MINUS(x, s(y))
MIN(s(u), s(v)) → MIN(u, v)
EQUAL(s(x), s(y)) → EQUAL(x, y)

The TRS R consists of the following rules:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(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 3 SCCs with 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

EQUAL(s(x), s(y)) → EQUAL(x, y)

The TRS R consists of the following rules:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(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:

EQUAL(s(x), s(y)) → EQUAL(x, y)

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

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(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].

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))

(11) Obligation:

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

EQUAL(s(x), s(y)) → EQUAL(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:

• EQUAL(s(x), s(y)) → EQUAL(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:

MIN(s(u), s(v)) → MIN(u, v)

The TRS R consists of the following rules:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(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:

MIN(s(u), s(v)) → MIN(u, v)

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

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(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)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))

(18) Obligation:

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

MIN(s(u), s(v)) → MIN(u, v)

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:

• MIN(s(u), s(v)) → MIN(u, v)
  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:

COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))

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:

COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))

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].

minus(x0, x1)
cond(true, x0, x1)

(25) Obligation:

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

COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))

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

(26) 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 COND(true, x, y) → MINUS(x, s(y)) the following chains were created:

• We consider the chain MINUS(x, y) → COND(equal(min(x, y), y), x, y), COND(true, x, y) → MINUS(x, s(y)) which results in the following constraint:
  

  (1)    (COND(equal(min(x2, x3), x3), x2, x3)=COND(true, x4, x5) ⇒ COND(true, x4, x5)≥MINUS(x4, s(x5)))

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

  (2)    (min(x2, x3)=x12∧equal(x12, x3)=true ⇒ COND(true, x2, x3)≥MINUS(x2, s(x3)))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on equal(x12, x3)=true which results in the following new constraints:
  

  (3)    (true=true∧min(x2, 0)=0 ⇒ COND(true, x2, 0)≥MINUS(x2, s(0)))

  
  

  (4)    (equal(x16, x15)=true∧min(x2, s(x15))=s(x16)∧(∀x17:equal(x16, x15)=true∧min(x17, x15)=x16 ⇒ COND(true, x17, x15)≥MINUS(x17, s(x15))) ⇒ COND(true, x2, s(x15))≥MINUS(x2, s(s(x15))))

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

  (5)    (0=x18∧min(x2, x18)=0 ⇒ COND(true, x2, 0)≥MINUS(x2, s(0)))

  
  
  We simplified constraint (4) using rule (VII) which results in the following new constraint:
  

  (6)    (equal(x16, x15)=true∧s(x15)=x23∧min(x2, x23)=s(x16)∧(∀x17:equal(x16, x15)=true∧min(x17, x15)=x16 ⇒ COND(true, x17, x15)≥MINUS(x17, s(x15))) ⇒ COND(true, x2, s(x15))≥MINUS(x2, s(s(x15))))

  
  
  We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on min(x2, x18)=0 which results in the following new constraints:
  

  (7)    (0=0∧0=x19 ⇒ COND(true, 0, 0)≥MINUS(0, s(0)))

  
  

  (8)    (0=0 ⇒ COND(true, x20, 0)≥MINUS(x20, s(0)))

  
  
  We simplified constraint (7) using rules (I), (II), (IV) which results in the following new constraint:
  

  (9)    (COND(true, 0, 0)≥MINUS(0, s(0)))

  
  
  We simplified constraint (8) using rules (I), (II) which results in the following new constraint:
  

  (10)    (COND(true, x20, 0)≥MINUS(x20, s(0)))

  
  
  We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on min(x2, x23)=s(x16) which results in the following new constraint:
  

  (11)    (s(min(x27, x26))=s(x16)∧equal(x16, x15)=true∧s(x15)=s(x26)∧(∀x17:equal(x16, x15)=true∧min(x17, x15)=x16 ⇒ COND(true, x17, x15)≥MINUS(x17, s(x15)))∧(∀x28,x29,x30:min(x27, x26)=s(x28)∧equal(x28, x29)=true∧s(x29)=x26∧(∀x30:equal(x28, x29)=true∧min(x30, x29)=x28 ⇒ COND(true, x30, x29)≥MINUS(x30, s(x29))) ⇒ COND(true, x27, s(x29))≥MINUS(x27, s(s(x29)))) ⇒ COND(true, s(x27), s(x15))≥MINUS(s(x27), s(s(x15))))

  
  
  We simplified constraint (11) using rules (I), (II), (IV) which results in the following new constraint:
  

  (12)    (min(x27, x26)=x16∧equal(x16, x15)=true∧x15=x26∧(∀x28,x29:min(x27, x26)=s(x28)∧equal(x28, x29)=true∧s(x29)=x26 ⇒ COND(true, x27, s(x29))≥MINUS(x27, s(s(x29)))) ⇒ COND(true, s(x27), s(x15))≥MINUS(s(x27), s(s(x15))))

  
  
  We simplified constraint (12) using rule (V) (with possible (I) afterwards) using induction on equal(x16, x15)=true which results in the following new constraints:
  

  (13)    (true=true∧min(x27, x26)=0∧0=x26∧(∀x28,x29:min(x27, x26)=s(x28)∧equal(x28, x29)=true∧s(x29)=x26 ⇒ COND(true, x27, s(x29))≥MINUS(x27, s(s(x29)))) ⇒ COND(true, s(x27), s(0))≥MINUS(s(x27), s(s(0))))

  
  

  (14)    (equal(x34, x33)=true∧min(x27, x26)=s(x34)∧s(x33)=x26∧(∀x28,x29:min(x27, x26)=s(x28)∧equal(x28, x29)=true∧s(x29)=x26 ⇒ COND(true, x27, s(x29))≥MINUS(x27, s(s(x29))))∧(∀x35,x36,x37,x38:equal(x34, x33)=true∧min(x35, x36)=x34∧x33=x36∧(∀x37,x38:min(x35, x36)=s(x37)∧equal(x37, x38)=true∧s(x38)=x36 ⇒ COND(true, x35, s(x38))≥MINUS(x35, s(s(x38)))) ⇒ COND(true, s(x35), s(x33))≥MINUS(s(x35), s(s(x33)))) ⇒ COND(true, s(x27), s(s(x33)))≥MINUS(s(x27), s(s(s(x33)))))

  
  
  We simplified constraint (13) using rules (I), (II), (IV) which results in the following new constraint:
  

  (15)    (min(x27, x26)=0∧0=x26 ⇒ COND(true, s(x27), s(0))≥MINUS(s(x27), s(s(0))))

  
  
  We simplified constraint (14) using rule (VI) where we applied the induction hypothesis (∀x28,x29:min(x27, x26)=s(x28)∧equal(x28, x29)=true∧s(x29)=x26 ⇒ COND(true, x27, s(x29))≥MINUS(x27, s(s(x29)))) with σ = [x28 / x34, x29 / x33] which results in the following new constraint:
  

  (16)    (COND(true, x27, s(x33))≥MINUS(x27, s(s(x33)))∧(∀x35,x36,x37,x38:equal(x34, x33)=true∧min(x35, x36)=x34∧x33=x36∧(∀x37,x38:min(x35, x36)=s(x37)∧equal(x37, x38)=true∧s(x38)=x36 ⇒ COND(true, x35, s(x38))≥MINUS(x35, s(s(x38)))) ⇒ COND(true, s(x35), s(x33))≥MINUS(s(x35), s(s(x33)))) ⇒ COND(true, s(x27), s(s(x33)))≥MINUS(s(x27), s(s(s(x33)))))

  
  
  We simplified constraint (15) using rule (V) (with possible (I) afterwards) using induction on min(x27, x26)=0 which results in the following new constraints:
  

  (17)    (0=0∧0=x39 ⇒ COND(true, s(0), s(0))≥MINUS(s(0), s(s(0))))

  
  

  (18)    (0=0 ⇒ COND(true, s(x40), s(0))≥MINUS(s(x40), s(s(0))))

  
  
  We simplified constraint (17) using rules (I), (II), (IV) which results in the following new constraint:
  

  (19)    (COND(true, s(0), s(0))≥MINUS(s(0), s(s(0))))

  
  
  We simplified constraint (18) using rules (I), (II) which results in the following new constraint:
  

  (20)    (COND(true, s(x40), s(0))≥MINUS(s(x40), s(s(0))))

  
  
  We simplified constraint (16) using rule (IV) which results in the following new constraint:
  

  (21)    (COND(true, x27, s(x33))≥MINUS(x27, s(s(x33))) ⇒ COND(true, s(x27), s(s(x33)))≥MINUS(s(x27), s(s(s(x33)))))

  
  
  

For Pair MINUS(x, y) → COND(equal(min(x, y), y), x, y) the following chains were created:

• We consider the chain COND(true, x, y) → MINUS(x, s(y)), MINUS(x, y) → COND(equal(min(x, y), y), x, y) which results in the following constraint:
  

  (22)    (MINUS(x6, s(x7))=MINUS(x8, x9) ⇒ MINUS(x8, x9)≥COND(equal(min(x8, x9), x9), x8, x9))

  
  
  We simplified constraint (22) using rules (I), (II), (III) which results in the following new constraint:
  

  (23)    (MINUS(x6, s(x7))≥COND(equal(min(x6, s(x7)), s(x7)), x6, s(x7)))

  
  
  

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

• COND(true, x, y) → MINUS(x, s(y))
  
  • (COND(true, 0, 0)≥MINUS(0, s(0)))
    
  • (COND(true, x20, 0)≥MINUS(x20, s(0)))
    
  • (COND(true, s(0), s(0))≥MINUS(s(0), s(s(0))))
    
  • (COND(true, s(x40), s(0))≥MINUS(s(x40), s(s(0))))
    
  • (COND(true, x27, s(x33))≥MINUS(x27, s(s(x33))) ⇒ COND(true, s(x27), s(s(x33)))≥MINUS(s(x27), s(s(s(x33)))))
    
  
  
• MINUS(x, y) → COND(equal(min(x, y), y), x, y)
  
  • (MINUS(x6, s(x7))≥COND(equal(min(x6, s(x7)), s(x7)), x6, s(x7)))
    
  
  

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(COND(x₁, x₂, x₃)) = -1 - x₁ + x₂ - x₃   
POL(MINUS(x₁, x₂)) = -1 + x₁ - x₂   
POL(c) = -2   
POL(equal(x₁, x₂)) = 0   
POL(false) = 0   
POL(min(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

COND(true, x, y) → MINUS(x, s(y))

The following pairs are in P_bound:

COND(true, x, y) → MINUS(x, s(y))

The following rules are usable:

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

(27) Obligation:

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

MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(x0), s(x1))

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

(28) DependencyGraphProof (EQUIVALENT transformation)

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

(29) TRUE