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

0 QTRS
↳1 AAECC Innermost (⇔)
↳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 QDPSizeChangeProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 QReductionProof (⇔)
    ↳25 QDP
    ↳26 Induction-Processor (⇐)
    ↳27 AND
      ↳28 QDP
        ↳29 PisEmptyProof (⇔)
        ↳30 TRUE
      ↳31 QTRS
        ↳32 QTRSRRRProof (⇔)
        ↳33 QTRS
        ↳34 QTRSRRRProof (⇔)
        ↳35 QTRS
        ↳36 QTRSRRRProof (⇔)
        ↳37 QTRS
        ↳38 QTRSRRRProof (⇔)
        ↳39 QTRS
        ↳40 QTRSRRRProof (⇔)
        ↳41 QTRS
          ↳42 RisEmptyProof (⇔)
            ↳43 TRUE
          ↳44 RisEmptyProof (⇔)
            ↳45 TRUE

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

The TRS R 2 is

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

The signature Sigma is {le, true, false}

(2) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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(s(x), s(y)) → MINUS(x, y)
LE(s(x), s(y)) → LE(x, y)
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))
QUOT(s(x), s(y)) → MINUS(s(x), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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:

LE(s(x), s(y)) → LE(x, y)

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

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

(11) Obligation:

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

LE(s(x), s(y)) → LE(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:

  • LE(s(x), s(y)) → LE(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(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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:

MINUS(s(x), s(y)) → MINUS(x, y)

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

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

(18) Obligation:

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

MINUS(s(x), s(y)) → MINUS(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:

  • MINUS(s(x), s(y)) → MINUS(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:

QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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:

QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

(25) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

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

(26) Induction-Processor (SOUND transformation)


This DP could be deleted by the Induction-Processor:
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(QUOT(x1, x2)) = x1   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

At least one of these decreasing rules is always used after the deleted DP:
minus(s(x), s(y)) → minus(x, y)


The following formula is valid:
x:sort[a0],y:sort[a0].minus'(s(x ), s(y ))=true


The transformed set:
minus'(s(x), s(y)) → true
minus'(x1, 0) → false
minus'(0, s(x0)) → false
minus'(s(x2), 0) → false
minus(s(x), s(y)) → minus(x, y)
minus(x1, 0) → x1
minus(0, s(x0)) → 0
minus(s(x2), 0) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a11](witness_sort[a11], witness_sort[a11]) → true

(27) Complex Obligation (AND)

(28) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

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

(29) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(30) TRUE

(31) Obligation:

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

minus'(s(x), s(y)) → true
minus'(x1, 0) → false
minus'(0, s(x0)) → false
minus'(s(x2), 0) → false
minus(s(x), s(y)) → minus(x, y)
minus(x1, 0) → x1
minus(0, s(x0)) → 0
minus(s(x2), 0) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a11](witness_sort[a11], witness_sort[a11]) → true

Q is empty.

(32) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(equal_bool(x1, x2)) = x1 + x2   
POL(equal_sort[a0](x1, x2)) = 1 + x1 + x2   
POL(equal_sort[a11](x1, x2)) = 1 + x1 + x2   
POL(false) = 2   
POL(isa_false(x1)) = 3 + x1   
POL(isa_true(x1)) = x1   
POL(minus(x1, x2)) = x1 + x2   
POL(minus'(x1, x2)) = 1 + x1 + x2   
POL(not(x1)) = 3 + x1   
POL(or(x1, x2)) = 1 + x1 + x2   
POL(s(x1)) = x1   
POL(true) = 0   
POL(witness_sort[a11]) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

minus'(s(x), s(y)) → true
minus(x1, 0) → x1
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a11](witness_sort[a11], witness_sort[a11]) → true


(33) Obligation:

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

minus'(x1, 0) → false
minus'(0, s(x0)) → false
minus'(s(x2), 0) → false
minus(s(x), s(y)) → minus(x, y)
minus(0, s(x0)) → 0
minus(s(x2), 0) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
isa_true(true) → true
isa_true(false) → false
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

Q is empty.

(34) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(equal_bool(x1, x2)) = x1 + x2   
POL(equal_sort[a0](x1, x2)) = x1 + x2   
POL(false) = 0   
POL(isa_true(x1)) = x1   
POL(minus(x1, x2)) = x1 + x2   
POL(minus'(x1, x2)) = 1 + x1 + x2   
POL(s(x1)) = x1   
POL(true) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

minus'(x1, 0) → false
minus'(0, s(x0)) → false
minus'(s(x2), 0) → false


(35) Obligation:

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

minus(s(x), s(y)) → minus(x, y)
minus(0, s(x0)) → 0
minus(s(x2), 0) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
isa_true(true) → true
isa_true(false) → false
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

Q is empty.

(36) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(equal_bool(x1, x2)) = x1 + x2   
POL(equal_sort[a0](x1, x2)) = x1 + x2   
POL(false) = 0   
POL(isa_true(x1)) = x1   
POL(minus(x1, x2)) = 1 + x1 + x2   
POL(s(x1)) = x1   
POL(true) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

minus(0, s(x0)) → 0
minus(s(x2), 0) → 0


(37) Obligation:

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

minus(s(x), s(y)) → minus(x, y)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
isa_true(true) → true
isa_true(false) → false
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

Q is empty.

(38) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(equal_bool(x1, x2)) = 1 + x1 + x2   
POL(equal_sort[a0](x1, x2)) = x1 + x2   
POL(false) = 0   
POL(isa_true(x1)) = x1   
POL(minus(x1, x2)) = x1 + x2   
POL(s(x1)) = x1   
POL(true) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true


(39) Obligation:

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

minus(s(x), s(y)) → minus(x, y)
isa_true(true) → true
isa_true(false) → false
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

Q is empty.

(40) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 2   
POL(equal_sort[a0](x1, x2)) = 1 + x1 + x2   
POL(false) = 1   
POL(isa_true(x1)) = 2 + 2·x1   
POL(minus(x1, x2)) = 2·x1 + x2   
POL(s(x1)) = 1 + 2·x1   
POL(true) = 1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

minus(s(x), s(y)) → minus(x, y)
isa_true(true) → true
isa_true(false) → false
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)


(41) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(42) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(43) TRUE

(44) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(45) TRUE