(0) Obligation:

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

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

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:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, 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:

LE(s(x), s(y)) → LE(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) → LE(y, x)
IF_MOD(true, x, y) → MOD(minus(x, y), y)
IF_MOD(true, x, y) → MINUS(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, 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:

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, 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:

MINUS(s(x), s(y)) → MINUS(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, 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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(x0), s(x1))

(11) 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.

(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:

  • MINUS(s(x), s(y)) → MINUS(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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, 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:

LE(s(x), s(y)) → LE(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, 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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(x0), s(x1))

(18) 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.

(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:

  • LE(s(x), s(y)) → LE(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:

MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
IF_MOD(true, x, y) → MOD(minus(x, y), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, 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:

MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
IF_MOD(true, x, y) → MOD(minus(x, y), 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)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, 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].

mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(x0), s(x1))

(25) Obligation:

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

MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
IF_MOD(true, x, y) → MOD(minus(x, y), 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)

The set Q consists of the following terms:

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

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

(26) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF_MOD(true, x, y) → MOD(minus(x, y), y) we obtained the following new rules [LPAR04]:

IF_MOD(true, s(z0), s(z1)) → MOD(minus(s(z0), s(z1)), s(z1))

(27) Obligation:

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

MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
IF_MOD(true, s(z0), s(z1)) → MOD(minus(s(z0), s(z1)), s(z1))

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)

The set Q consists of the following terms:

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

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

(28) Induction-Processor (SOUND transformation)


This DP could be deleted by the Induction-Processor:
IF_MOD(true, s(z0), s(z1)) → MOD(minus(s(z0), s(z1)), s(z1))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(IF_MOD(x1, x2, x3)) = x2   
POL(MOD(x1, x2)) = x1   
POL(false) = 0   
POL(le(x1, x2)) = x2   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

At least one of these decreasing rules is always used after the deleted DP:
minus(s(x3'), s(y1')) → minus(x3', y1')


The following formula is valid:
z2:sort[a0],z3:sort[a0].((¬(z2 =0)∧¬(z3 =0))→minus'(z2 , z3 )=true)


The transformed set:
minus'(x, 0) → false
minus'(s(x3), s(y1)) → true
minus'(0, s(x0)) → false
minus(x, 0) → x
minus(s(x3), s(y1)) → minus(x3, y1)
le(0, y5) → true
le(s(x12), 0) → false
le(s(x17), s(y12)) → le(x17, y12)
minus(0, s(x0)) → 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[a23](witness_sort[a23], witness_sort[a23]) → true

(29) Complex Obligation (AND)

(30) Obligation:

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

MOD(s(x), s(y)) → IF_MOD(le(y, x), 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)

The set Q consists of the following terms:

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

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

(31) DependencyGraphProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

minus'(x, 0) → false
minus'(s(x3), s(y1)) → true
minus'(0, s(x0)) → false
minus(x, 0) → x
minus(s(x3), s(y1)) → minus(x3, y1)
le(0, y5) → true
le(s(x12), 0) → false
le(s(x17), s(y12)) → le(x17, y12)
minus(0, s(x0)) → 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[a23](witness_sort[a23], witness_sort[a23]) → true

Q is empty.

(34) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
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[a23](x1, x2)) = 1 + x1 + x2   
POL(false) = 1   
POL(isa_false(x1)) = 2 + x1   
POL(isa_true(x1)) = x1   
POL(le(x1, x2)) = 2 + x1 + x2   
POL(minus(x1, x2)) = 1 + x1 + x2   
POL(minus'(x1, x2)) = 1 + x1 + x2   
POL(not(x1)) = 2 + x1   
POL(or(x1, x2)) = x1 + x2   
POL(s(x1)) = x1   
POL(true) = 0   
POL(witness_sort[a23]) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

minus'(s(x3), s(y1)) → true
minus(x, 0) → x
le(0, y5) → true
le(s(x12), 0) → false
minus(0, s(x0)) → 0
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
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[a23](witness_sort[a23], witness_sort[a23]) → true


(35) Obligation:

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

minus'(x, 0) → false
minus'(0, s(x0)) → false
minus(s(x3), s(y1)) → minus(x3, y1)
le(s(x17), s(y12)) → le(x17, y12)
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
or(true, x) → 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(le(x1, x2)) = x1 + x2   
POL(minus(x1, x2)) = x1 + x2   
POL(minus'(x1, x2)) = 1 + x1 + x2   
POL(or(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'(x, 0) → false
minus'(0, s(x0)) → false
or(true, x) → true


(37) Obligation:

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

minus(s(x3), s(y1)) → minus(x3, y1)
le(s(x17), s(y12)) → le(x17, y12)
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(le(x1, x2)) = x1 + x2   
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(x3), s(y1)) → minus(x3, y1)
le(s(x17), s(y12)) → le(x17, y12)
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) = 0   
POL(equal_sort[a0](x1, x2)) = x1 + x2   
POL(false) = 0   
POL(isa_true(x1)) = 1 + x1   
POL(le(x1, x2)) = x1 + x2   
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:

isa_true(true) → true
isa_true(false) → false


(41) Obligation:

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

minus(s(x3), s(y1)) → minus(x3, y1)
le(s(x17), s(y12)) → le(x17, y12)
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.

(42) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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

minus(s(x3), s(y1)) → minus(x3, y1)
le(s(x17), s(y12)) → le(x17, y12)
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)


(43) Obligation:

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

(44) RisEmptyProof (EQUIVALENT transformation)

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

(45) TRUE

(46) RisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) RisEmptyProof (EQUIVALENT transformation)

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

(49) TRUE