Termination w.r.t. Q proof of GTSSK07

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

0 QTRS

↳1 Overlay + Local Confluence (⇔, 0 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 7 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 QReductionProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 QDPSizeChangeProof (⇔, 0 ms)

        ↳13 YES

    ↳14 QDP

        ↳15 UsableRulesProof (⇔, 0 ms)

        ↳16 QDP

        ↳17 QReductionProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 QDPSizeChangeProof (⇔, 0 ms)

        ↳20 YES

    ↳21 QDP

        ↳22 UsableRulesProof (⇔, 0 ms)

        ↳23 QDP

        ↳24 QReductionProof (⇔, 0 ms)

        ↳25 QDP

        ↳26 QDPQMonotonicMRRProof (⇔, 0 ms)

        ↳27 QDP

        ↳28 NonInfProof (⇔, 0 ms)

        ↳29 QDP

        ↳30 DependencyGraphProof (⇔, 0 ms)

        ↳31 TRUE

(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) YES

(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) YES

(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) QDPQMonotonicMRRProof (EQUIVALENT transformation)

By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

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

(27) 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), 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) 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(x2, x3) → COND(equal(min(x2, x3), x3), x2, x3), COND(true, x4, x5) → MINUS(x4, s(x5)) 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(x14, x13)=true∧min(x2, s(x13))=s(x14)∧(∀x15:equal(x14, x13)=true∧min(x15, x13)=x14 ⇒ COND(true, x15, x13)≥MINUS(x15, s(x13))) ⇒ COND(true, x2, s(x13))≥MINUS(x2, s(s(x13))))

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

(5) (0=x16∧min(x2, x16)=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(x14, x13)=true∧s(x13)=x21∧min(x2, x21)=s(x14)∧(∀x15:equal(x14, x13)=true∧min(x15, x13)=x14 ⇒ COND(true, x15, x13)≥MINUS(x15, s(x13))) ⇒ COND(true, x2, s(x13))≥MINUS(x2, s(s(x13))))

We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on min(x2, x16)=0 which results in the following new constraints:

(7) (0=0∧0=x17 ⇒ COND(true, 0, 0)≥MINUS(0, s(0)))

(8) (0=0 ⇒ COND(true, x18, 0)≥MINUS(x18, 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, x18, 0)≥MINUS(x18, s(0)))

We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on min(x2, x21)=s(x14) which results in the following new constraint:

(11) (s(min(x25, x24))=s(x14)∧equal(x14, x13)=true∧s(x13)=s(x24)∧(∀x15:equal(x14, x13)=true∧min(x15, x13)=x14 ⇒ COND(true, x15, x13)≥MINUS(x15, s(x13)))∧(∀x26,x27,x28:min(x25, x24)=s(x26)∧equal(x26, x27)=true∧s(x27)=x24∧(∀x28:equal(x26, x27)=true∧min(x28, x27)=x26 ⇒ COND(true, x28, x27)≥MINUS(x28, s(x27))) ⇒ COND(true, x25, s(x27))≥MINUS(x25, s(s(x27)))) ⇒ COND(true, s(x25), s(x13))≥MINUS(s(x25), s(s(x13))))

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

(12) (min(x25, x24)=x14∧equal(x14, x13)=true∧x13=x24 ⇒ COND(true, s(x25), s(x13))≥MINUS(s(x25), s(s(x13))))

We simplified constraint (12) using rule (V) (with possible (I) afterwards) using induction on equal(x14, x13)=true which results in the following new constraints:

(13) (true=true∧min(x25, x24)=0∧0=x24 ⇒ COND(true, s(x25), s(0))≥MINUS(s(x25), s(s(0))))

(14) (equal(x30, x29)=true∧min(x25, x24)=s(x30)∧s(x29)=x24∧(∀x31,x32:equal(x30, x29)=true∧min(x31, x32)=x30∧x29=x32 ⇒ COND(true, s(x31), s(x29))≥MINUS(s(x31), s(s(x29)))) ⇒ COND(true, s(x25), s(s(x29)))≥MINUS(s(x25), s(s(s(x29)))))

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

(15) (min(x25, x24)=0∧0=x24 ⇒ COND(true, s(x25), s(0))≥MINUS(s(x25), s(s(0))))

We simplified constraint (14) using rule (V) (with possible (I) afterwards) using induction on min(x25, x24)=s(x30) which results in the following new constraint:

(16) (s(min(x40, x39))=s(x30)∧equal(x30, x29)=true∧s(x29)=s(x39)∧(∀x31,x32:equal(x30, x29)=true∧min(x31, x32)=x30∧x29=x32 ⇒ COND(true, s(x31), s(x29))≥MINUS(s(x31), s(s(x29))))∧(∀x41,x42,x43,x44:min(x40, x39)=s(x41)∧equal(x41, x42)=true∧s(x42)=x39∧(∀x43,x44:equal(x41, x42)=true∧min(x43, x44)=x41∧x42=x44 ⇒ COND(true, s(x43), s(x42))≥MINUS(s(x43), s(s(x42)))) ⇒ COND(true, s(x40), s(s(x42)))≥MINUS(s(x40), s(s(s(x42))))) ⇒ COND(true, s(s(x40)), s(s(x29)))≥MINUS(s(s(x40)), s(s(s(x29)))))

We simplified constraint (15) using rule (V) (with possible (I) afterwards) using induction on min(x25, x24)=0 which results in the following new constraints:

(17) (0=0∧0=x33 ⇒ COND(true, s(0), s(0))≥MINUS(s(0), s(s(0))))

(18) (0=0 ⇒ COND(true, s(x34), s(0))≥MINUS(s(x34), 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(x34), s(0))≥MINUS(s(x34), s(s(0))))

We simplified constraint (16) using rules (I), (II) which results in the following new constraint:

(21) (min(x40, x39)=x30∧equal(x30, x29)=true∧x29=x39∧(∀x31,x32:equal(x30, x29)=true∧min(x31, x32)=x30∧x29=x32 ⇒ COND(true, s(x31), s(x29))≥MINUS(s(x31), s(s(x29))))∧(∀x41,x42,x43,x44:min(x40, x39)=s(x41)∧equal(x41, x42)=true∧s(x42)=x39∧(∀x43,x44:equal(x41, x42)=true∧min(x43, x44)=x41∧x42=x44 ⇒ COND(true, s(x43), s(x42))≥MINUS(s(x43), s(s(x42)))) ⇒ COND(true, s(x40), s(s(x42)))≥MINUS(s(x40), s(s(s(x42))))) ⇒ COND(true, s(s(x40)), s(s(x29)))≥MINUS(s(s(x40)), s(s(s(x29)))))

We simplified constraint (21) using rule (VI) where we applied the induction hypothesis (∀x31,x32:equal(x30, x29)=true∧min(x31, x32)=x30∧x29=x32 ⇒ COND(true, s(x31), s(x29))≥MINUS(s(x31), s(s(x29)))) with σ = [x31 / x40, x32 / x39] which results in the following new constraint:

(22) (COND(true, s(x40), s(x29))≥MINUS(s(x40), s(s(x29)))∧(∀x41,x42,x43,x44:min(x40, x39)=s(x41)∧equal(x41, x42)=true∧s(x42)=x39∧(∀x43,x44:equal(x41, x42)=true∧min(x43, x44)=x41∧x42=x44 ⇒ COND(true, s(x43), s(x42))≥MINUS(s(x43), s(s(x42)))) ⇒ COND(true, s(x40), s(s(x42)))≥MINUS(s(x40), s(s(s(x42))))) ⇒ COND(true, s(s(x40)), s(s(x29)))≥MINUS(s(s(x40)), s(s(s(x29)))))

We simplified constraint (22) using rule (IV) which results in the following new constraint:

(23) (COND(true, s(x40), s(x29))≥MINUS(s(x40), s(s(x29))) ⇒ COND(true, s(s(x40)), s(s(x29)))≥MINUS(s(s(x40)), s(s(s(x29)))))

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

We consider the chain COND(true, x6, x7) → MINUS(x6, s(x7)), MINUS(x8, x9) → COND(equal(min(x8, x9), x9), x8, x9) which results in the following constraint:
(1) (MINUS(x6, s(x7))=MINUS(x8, x9) ⇒ MINUS(x8, x9)≥COND(equal(min(x8, x9), x9), x8, x9))

We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2) (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, x18, 0)≥MINUS(x18, s(0)))
- (COND(true, s(0), s(0))≥MINUS(s(0), s(s(0))))
- (COND(true, s(x34), s(0))≥MINUS(s(x34), s(s(0))))
- (COND(true, s(x40), s(x29))≥MINUS(s(x40), s(s(x29))) ⇒ COND(true, s(s(x40)), s(s(x29)))≥MINUS(s(s(x40)), s(s(s(x29)))))
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)) = x₁ + x₂ - x₃
POL(MINUS(x₁, x₂)) = x₁ - x₂
POL(c) = -1
POL(equal(x₁, x₂)) = 0
POL(min(x₁, x₂)) = x₂
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:

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

(29) 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), 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.

(30) DependencyGraphProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) Overlay + Local Confluence (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) UsableRulesProof (EQUIVALENT transformation)

(9) Obligation:

(10) QReductionProof (EQUIVALENT transformation)

(11) Obligation:

(12) QDPSizeChangeProof (EQUIVALENT transformation)

(13) YES

(14) Obligation:

(15) UsableRulesProof (EQUIVALENT transformation)

(16) Obligation:

(17) QReductionProof (EQUIVALENT transformation)

(18) Obligation:

(19) QDPSizeChangeProof (EQUIVALENT transformation)

(20) YES

(21) Obligation:

(22) UsableRulesProof (EQUIVALENT transformation)

(23) Obligation:

(24) QReductionProof (EQUIVALENT transformation)

(25) Obligation:

(26) QDPQMonotonicMRRProof (EQUIVALENT transformation)

(27) Obligation:

(28) NonInfProof (EQUIVALENT transformation)

(29) Obligation:

(30) DependencyGraphProof (EQUIVALENT transformation)

(31) TRUE