YES Termination w.r.t. Q proof of Strategy_removed_AG01_4.30b.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 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 TransformationProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 Induction-Processor (⇒, 7 ms)
        ↳29 AND
            ↳30 QDP
                ↳31 DependencyGraphProof (⇔, 0 ms)
                ↳32 TRUE
            ↳33 QTRS
                ↳34 QTRSRRRProof (⇔, 0 ms)
                ↳35 QTRS
                ↳36 RisEmptyProof (⇔, 0 ms)
                ↳37 YES

(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) YES

(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) YES

(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) TransformationProof (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)) → 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_renamed, s(z0), s(z1)) → MOD(minus(s(z0), s(z1)), s(z1))


This order was computed:
Polynomial interpretation [POLO]:

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

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:
z2:sort[a0],z3:sort[a0].((¬(z2 =0)∧¬(z3 =0))→minus'(z2 , z3 )=true)


The transformed set:
minus'(x', 0) → false
minus'(s(x''), s(y')) → true
minus'(0, s(v11)) → false
minus(x', 0) → x'
minus(s(x''), s(y')) → minus(x'', y')
le(0, y'') → true_renamed
le(s(x1), 0) → false_renamed
le(s(x2), s(y1)) → le(x2, y1)
minus(0, s(v11)) → 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(v18)) → false
equal_sort[a0](s(v19), 0) → false
equal_sort[a0](s(v19), s(v20)) → equal_sort[a0](v19, v20)
equal_sort[a19](true_renamed, true_renamed) → true
equal_sort[a19](true_renamed, false_renamed) → false
equal_sort[a19](false_renamed, true_renamed) → false
equal_sort[a19](false_renamed, false_renamed) → true
equal_sort[a24](witness_sort[a24], witness_sort[a24]) → true


The proof given by the theorem prover:
The following output was given by the internal theorem prover:
proof of internal
# AProVE Commit ID: 3a20a6ef7432c3f292db1a8838479c42bf5e3b22 root 20240618 unpublished


Partial correctness of the following Program

   [x, v18, v19, v20, x', x'', y', v11, y'', x1, x2, y1]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or 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(v18)) -> false
   equal_sort[a0](s(v19), 0) -> false
   equal_sort[a0](s(v19), s(v20)) -> equal_sort[a0](v19, v20)
   equal_sort[a19](true_renamed, true_renamed) -> true
   equal_sort[a19](true_renamed, false_renamed) -> false
   equal_sort[a19](false_renamed, true_renamed) -> false
   equal_sort[a19](false_renamed, false_renamed) -> true
   equal_sort[a24](witness_sort[a24], witness_sort[a24]) -> true
   minus'(x', 0) -> false
   minus'(s(x''), s(y')) -> true
   minus'(0, s(v11)) -> false
   minus(x', 0) -> x'
   minus(s(x''), s(y')) -> minus(x'', y')
   minus(0, s(v11)) -> 0
   le(0, y'') -> true_renamed
   le(s(x1), 0) -> false_renamed
   le(s(x2), s(y1)) -> le(x2, y1)

using the following formula:
z2:sort[a0],z3:sort[a0].((~(z2=0)/\~(z3=0))->minus'(z2, z3)=true)

could be successfully shown:
(0) Formula
(1) Induction by data structure [EQUIVALENT, 0 ms]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT, 0 ms]
        (5) YES
    (6) Formula
        (7) Symbolic evaluation [EQUIVALENT, 0 ms]
        (8) Formula
        (9) Case Analysis [EQUIVALENT, 0 ms]
        (10) AND
            (11) Formula
                (12) Symbolic evaluation [EQUIVALENT, 0 ms]
                (13) YES
            (14) Formula
                (15) Symbolic evaluation [EQUIVALENT, 0 ms]
                (16) YES


----------------------------------------

(0)
Obligation:
Formula:
z2:sort[a0],z3:sort[a0].((~(z2=0)/\~(z3=0))->minus'(z2, z3)=true)

There are no hypotheses.




----------------------------------------

(1) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
z3:sort[a0].((~(0=0)/\~(z3=0))->minus'(0, z3)=true)

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0],z3:sort[a0].((~(s(n)=0)/\~(z3=0))->minus'(s(n), z3)=true)

Hypotheses:
n:sort[a0],!z3:sort[a0].((~(n=0)/\~(z3=0))->minus'(n, z3)=true)






----------------------------------------

(2)
Complex Obligation (AND)

----------------------------------------

(3)
Obligation:
Formula:
z3:sort[a0].((~(0=0)/\~(z3=0))->minus'(0, z3)=true)

There are no hypotheses.




----------------------------------------

(4) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(5)
YES

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(6)
Obligation:
Formula:
n:sort[a0],z3:sort[a0].((~(s(n)=0)/\~(z3=0))->minus'(s(n), z3)=true)

Hypotheses:
n:sort[a0],!z3:sort[a0].((~(n=0)/\~(z3=0))->minus'(n, z3)=true)




----------------------------------------

(7) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(8)
Obligation:
Formula:
z3:sort[a0],n:sort[a0].(~(z3=0)->minus'(s(n), z3)=true)

Hypotheses:
n:sort[a0],!z3:sort[a0].((~(n=0)/\~(z3=0))->minus'(n, z3)=true)




----------------------------------------

(9) Case Analysis (EQUIVALENT)
Case analysis leads to the following new obligations:

Formula:
n:sort[a0].(~(0=0)->minus'(s(n), 0)=true)

Hypotheses:
n:sort[a0],!z3:sort[a0].((~(n=0)/\~(z3=0))->minus'(n, z3)=true)





Formula:
x_1:sort[a0],n:sort[a0].(~(s(x_1)=0)->minus'(s(n), s(x_1))=true)

Hypotheses:
n:sort[a0],!z3:sort[a0].((~(n=0)/\~(z3=0))->minus'(n, z3)=true)






----------------------------------------

(10)
Complex Obligation (AND)

----------------------------------------

(11)
Obligation:
Formula:
n:sort[a0].(~(0=0)->minus'(s(n), 0)=true)

Hypotheses:
n:sort[a0],!z3:sort[a0].((~(n=0)/\~(z3=0))->minus'(n, z3)=true)




----------------------------------------

(12) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(13)
YES

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(14)
Obligation:
Formula:
x_1:sort[a0],n:sort[a0].(~(s(x_1)=0)->minus'(s(n), s(x_1))=true)

Hypotheses:
n:sort[a0],!z3:sort[a0].((~(n=0)/\~(z3=0))->minus'(n, z3)=true)




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(15) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(16)
YES

(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(x''), s(y')) → true
minus'(0, s(v11)) → false
minus(x', 0) → x'
minus(s(x''), s(y')) → minus(x'', y')
le(0, y'') → true_renamed
le(s(x1), 0) → false_renamed
le(s(x2), s(y1)) → le(x2, y1)
minus(0, s(v11)) → 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(v18)) → false
equal_sort[a0](s(v19), 0) → false
equal_sort[a0](s(v19), s(v20)) → equal_sort[a0](v19, v20)
equal_sort[a19](true_renamed, true_renamed) → true
equal_sort[a19](true_renamed, false_renamed) → false
equal_sort[a19](false_renamed, true_renamed) → false
equal_sort[a19](false_renamed, false_renamed) → true
equal_sort[a24](witness_sort[a24], witness_sort[a24]) → true

Q is empty.

(34) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Knuth-Bendix order [KBO] with precedence:
isatrue1 > witnesssort[a24] > equalsort[a24]2 > equalsort[a19]2 > or2 > le2 > equalsort[a0]2 > not1 > isafalse1 > equalbool2 > minus2 > 0 > false > minus'2 > true > s1 > truerenamed > and2 > falserenamed

and weight map:

0=4
false=9
true=9
true_renamed=5
false_renamed=6
witness_sort[a24]=1
s_1=1
not_1=1
isa_true_1=0
isa_false_1=1
minus'_2=5
minus_2=0
le_2=0
equal_bool_2=0
and_2=0
or_2=0
equal_sort[a0]_2=3
equal_sort[a19]_2=0
equal_sort[a24]_2=7

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

minus'(x', 0) → false
minus'(s(x''), s(y')) → true
minus'(0, s(v11)) → false
minus(x', 0) → x'
minus(s(x''), s(y')) → minus(x'', y')
le(0, y'') → true_renamed
le(s(x1), 0) → false_renamed
le(s(x2), s(y1)) → le(x2, y1)
minus(0, s(v11)) → 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(v18)) → false
equal_sort[a0](s(v19), 0) → false
equal_sort[a0](s(v19), s(v20)) → equal_sort[a0](v19, v20)
equal_sort[a19](true_renamed, true_renamed) → true
equal_sort[a19](true_renamed, false_renamed) → false
equal_sort[a19](false_renamed, true_renamed) → false
equal_sort[a19](false_renamed, false_renamed) → true
equal_sort[a24](witness_sort[a24], witness_sort[a24]) → true


(35) Obligation:

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

(36) RisEmptyProof (EQUIVALENT transformation)

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

(37) YES