Termination Proof using AProVE

Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(x, s(y)) -> PRED(minus(x, y))
MINUS(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, s(x), s(y)) -> MOD(minus(x, y), s(y))
IF_MOD(true, s(x), s(y)) -> MINUS(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

LE(s(x), s(y)) -> LE(x, y)

one new Dependency Pair is created:

LE(s(s(x'')), s(s(y''))) -> LE(s(x''), s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

LE(s(s(x'')), s(s(y''))) -> LE(s(x''), s(y''))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

LE(s(s(x'')), s(s(y''))) -> LE(s(x''), s(y''))

one new Dependency Pair is created:

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(s(s(x'''')), s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(s(s(x'''')), s(s(y'''')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(s(s(x'''')), s(s(y'''')))

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

LE(x₁, x₂) -> LE(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(x, s(y)) -> MINUS(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

MINUS(x, s(y)) -> MINUS(x, y)

one new Dependency Pair is created:

MINUS(x'', s(s(y''))) -> MINUS(x'', s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(x'', s(s(y''))) -> MINUS(x'', s(y''))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

MINUS(x'', s(s(y''))) -> MINUS(x'', s(y''))

one new Dependency Pair is created:

MINUS(x'''', s(s(s(y'''')))) -> MINUS(x'''', s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(x'''', s(s(s(y'''')))) -> MINUS(x'''', s(s(y'''')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(x'''', s(s(s(y'''')))) -> MINUS(x'''', s(s(y'''')))

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pairs:

IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
MOD(s(x), s(y)) -> IF_MOD(le(y, x), s(x), s(y))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MOD(s(x), s(y)) -> IF_MOD(le(y, x), s(x), s(y))

three new Dependency Pairs are created:

MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))
MOD(s(0), s(s(x''))) -> IF_MOD(false, s(0), s(s(x'')))
MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Narrowing Transformation

Dependency Pairs:

MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))
MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))
IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))

two new Dependency Pairs are created:

IF_MOD(true, s(x''), s(0)) -> MOD(x'', s(0))
IF_MOD(true, s(x''), s(s(y''))) -> MOD(pred(minus(x'', y'')), s(s(y'')))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 11

                 ↳Narrowing Transformation

Dependency Pairs:

IF_MOD(true, s(x''), s(s(y''))) -> MOD(pred(minus(x'', y'')), s(s(y'')))
MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))

three new Dependency Pairs are created:

MOD(s(s(y''')), s(s(0))) -> IF_MOD(true, s(s(y''')), s(s(0)))
MOD(s(s(0)), s(s(s(x')))) -> IF_MOD(false, s(s(0)), s(s(s(x'))))
MOD(s(s(s(y'))), s(s(s(x')))) -> IF_MOD(le(x', y'), s(s(s(y'))), s(s(s(x'))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 13

                 ↳Narrowing Transformation

Dependency Pairs:

MOD(s(s(s(y'))), s(s(s(x')))) -> IF_MOD(le(x', y'), s(s(s(y'))), s(s(s(x'))))
MOD(s(s(y''')), s(s(0))) -> IF_MOD(true, s(s(y''')), s(s(0)))
IF_MOD(true, s(x''), s(s(y''))) -> MOD(pred(minus(x'', y'')), s(s(y'')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF_MOD(true, s(x''), s(s(y''))) -> MOD(pred(minus(x'', y'')), s(s(y'')))

two new Dependency Pairs are created:

IF_MOD(true, s(x'''), s(s(0))) -> MOD(pred(x'''), s(s(0)))
IF_MOD(true, s(x'''), s(s(s(y')))) -> MOD(pred(pred(minus(x''', y'))), s(s(s(y'))))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 15

                 ↳Argument Filtering and Ordering

Dependency Pairs:

IF_MOD(true, s(x'''), s(s(s(y')))) -> MOD(pred(pred(minus(x''', y'))), s(s(s(y'))))
MOD(s(s(s(y'))), s(s(s(x')))) -> IF_MOD(le(x', y'), s(s(s(y'))), s(s(s(x'))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MOD(true, s(x'''), s(s(s(y')))) -> MOD(pred(pred(minus(x''', y'))), s(s(s(y'))))

The following usable rules for innermost can be oriented:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

pred > true
{MOD, IF_MOD} > true
0 > false > true
minus > true
s > true
le > true

resulting in one new DP problem.
Used Argument Filtering System:

MOD(x₁, x₂) -> MOD(x₁, x₂)
IF_MOD(x₁, x₂, x₃) -> IF_MOD(x₂, x₃)
s(x₁) -> s(x₁)
pred(x₁) -> x₁
minus(x₁, x₂) -> x₁
le(x₁, x₂) -> le(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 18

                 ↳Dependency Graph

Dependency Pair:

MOD(s(s(s(y'))), s(s(s(x')))) -> IF_MOD(le(x', y'), s(s(s(y'))), s(s(s(x'))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 16

                 ↳Argument Filtering and Ordering

Dependency Pairs:

IF_MOD(true, s(x'''), s(s(0))) -> MOD(pred(x'''), s(s(0)))
MOD(s(s(y''')), s(s(0))) -> IF_MOD(true, s(s(y''')), s(s(0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF_MOD(true, s(x'''), s(s(0))) -> MOD(pred(x'''), s(s(0)))
MOD(s(s(y''')), s(s(0))) -> IF_MOD(true, s(s(y''')), s(s(0)))

The following usable rule for innermost can be oriented:

pred(s(x)) -> x

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

s > pred > 0
s > {MOD, IF_MOD} > 0
s > true > 0

resulting in one new DP problem.
Used Argument Filtering System:

IF_MOD(x₁, x₂, x₃) -> IF_MOD(x₁, x₂, x₃)
MOD(x₁, x₂) -> MOD(x₁)
s(x₁) -> s(x₁)
pred(x₁) -> pred(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 19

                 ↳Dependency Graph

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 12

                 ↳Forward Instantiation Transformation

Dependency Pairs:

IF_MOD(true, s(x''), s(0)) -> MOD(x'', s(0))
MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_MOD(true, s(x''), s(0)) -> MOD(x'', s(0))

one new Dependency Pair is created:

IF_MOD(true, s(s(x'''')), s(0)) -> MOD(s(x''''), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

Dependency Pairs:

IF_MOD(true, s(s(x'''')), s(0)) -> MOD(s(x''''), s(0))
MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))

one new Dependency Pair is created:

MOD(s(s(x'''''')), s(0)) -> IF_MOD(true, s(s(x'''''')), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 17

                 ↳Argument Filtering and Ordering

Dependency Pairs:

MOD(s(s(x'''''')), s(0)) -> IF_MOD(true, s(s(x'''''')), s(0))
IF_MOD(true, s(s(x'''')), s(0)) -> MOD(s(x''''), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pairs can be strictly oriented:

MOD(s(s(x'''''')), s(0)) -> IF_MOD(true, s(s(x'''''')), s(0))
IF_MOD(true, s(s(x'''')), s(0)) -> MOD(s(x''''), s(0))

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

s > MOD > IF_MOD > true
s > 0 > true

resulting in one new DP problem.
Used Argument Filtering System:

MOD(x₁, x₂) -> MOD(x₁, x₂)
IF_MOD(x₁, x₂, x₃) -> IF_MOD(x₁, x₂, x₃)
s(x₁) -> s(x₁)

Innermost Termination of R successfully shown.
Duration:
0:15 minutes