YES

The TRS could be proven terminating. The proof took 1045 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (202ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (34ms).
 |    | – Problem 6 was processed with processor DependencyGraph (17ms).
 | – Problem 5 was processed with processor PolynomialLinearRange4iUR (394ms).
 |    | – Problem 7 was processed with processor PolynomialLinearRange4iUR (61ms).
 |    |    | – Problem 8 was processed with processor PolynomialLinearRange4iUR (69ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

inter#(app(l1, l2), l3)inter#(l2, l3)inter#(cons(x, l1), l2)ifinter#(mem(x, l2), x, l1, l2)
app#(app(l1, l2), l3)app#(l1, app(l2, l3))inter#(l1, app(l2, l3))app#(inter(l1, l2), inter(l1, l3))
inter#(l1, app(l2, l3))inter#(l1, l3)inter#(l1, cons(x, l2))mem#(x, l1)
app#(cons(x, l1), l2)app#(l1, l2)app#(app(l1, l2), l3)app#(l2, l3)
ifmem#(false, x, l)mem#(x, l)inter#(app(l1, l2), l3)app#(inter(l1, l3), inter(l2, l3))
inter#(l1, app(l2, l3))inter#(l1, l2)ifinter#(true, x, l1, l2)inter#(l1, l2)
mem#(x, cons(y, l))ifmem#(eq(x, y), x, l)inter#(cons(x, l1), l2)mem#(x, l2)
inter#(l1, cons(x, l2))ifinter#(mem(x, l1), x, l2, l1)ifinter#(false, x, l1, l2)inter#(l1, l2)
inter#(app(l1, l2), l3)inter#(l1, l3)mem#(x, cons(y, l))eq#(x, y)
eq#(s(x), s(y))eq#(x, y)

Rewrite Rules

if(true, x, y)xif(false, x, y)y
eq(0, 0)trueeq(0, s(x))false
eq(s(x), 0)falseeq(s(x), s(y))eq(x, y)
app(nil, l)lapp(cons(x, l1), l2)cons(x, app(l1, l2))
app(app(l1, l2), l3)app(l1, app(l2, l3))mem(x, nil)false
mem(x, cons(y, l))ifmem(eq(x, y), x, l)ifmem(true, x, l)true
ifmem(false, x, l)mem(x, l)inter(x, nil)nil
inter(nil, x)nilinter(app(l1, l2), l3)app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3))app(inter(l1, l2), inter(l1, l3))inter(cons(x, l1), l2)ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2))ifinter(mem(x, l1), x, l2, l1)ifinter(true, x, l1, l2)cons(x, inter(l1, l2))
ifinter(false, x, l1, l2)inter(l1, l2)

Original Signature

Termination of terms over the following signature is verified: app, ifmem, true, inter, 0, s, if, mem, false, ifinter, eq, nil, cons

Strategy


The following SCCs where found

app#(app(l1, l2), l3) → app#(l2, l3)app#(cons(x, l1), l2) → app#(l1, l2)
app#(app(l1, l2), l3) → app#(l1, app(l2, l3))

ifmem#(false, x, l) → mem#(x, l)mem#(x, cons(y, l)) → ifmem#(eq(x, y), x, l)

eq#(s(x), s(y)) → eq#(x, y)

inter#(l1, app(l2, l3)) → inter#(l1, l3)inter#(l1, app(l2, l3)) → inter#(l1, l2)
ifinter#(true, x, l1, l2) → inter#(l1, l2)inter#(l1, cons(x, l2)) → ifinter#(mem(x, l1), x, l2, l1)
inter#(app(l1, l2), l3) → inter#(l2, l3)ifinter#(false, x, l1, l2) → inter#(l1, l2)
inter#(cons(x, l1), l2) → ifinter#(mem(x, l2), x, l1, l2)inter#(app(l1, l2), l3) → inter#(l1, l3)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

eq#(s(x), s(y))eq#(x, y)

Rewrite Rules

if(true, x, y)xif(false, x, y)y
eq(0, 0)trueeq(0, s(x))false
eq(s(x), 0)falseeq(s(x), s(y))eq(x, y)
app(nil, l)lapp(cons(x, l1), l2)cons(x, app(l1, l2))
app(app(l1, l2), l3)app(l1, app(l2, l3))mem(x, nil)false
mem(x, cons(y, l))ifmem(eq(x, y), x, l)ifmem(true, x, l)true
ifmem(false, x, l)mem(x, l)inter(x, nil)nil
inter(nil, x)nilinter(app(l1, l2), l3)app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3))app(inter(l1, l2), inter(l1, l3))inter(cons(x, l1), l2)ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2))ifinter(mem(x, l1), x, l2, l1)ifinter(true, x, l1, l2)cons(x, inter(l1, l2))
ifinter(false, x, l1, l2)inter(l1, l2)

Original Signature

Termination of terms over the following signature is verified: app, ifmem, true, inter, 0, s, if, mem, false, ifinter, eq, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

eq#(s(x), s(y))eq#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

app#(app(l1, l2), l3)app#(l2, l3)app#(cons(x, l1), l2)app#(l1, l2)
app#(app(l1, l2), l3)app#(l1, app(l2, l3))

Rewrite Rules

if(true, x, y)xif(false, x, y)y
eq(0, 0)trueeq(0, s(x))false
eq(s(x), 0)falseeq(s(x), s(y))eq(x, y)
app(nil, l)lapp(cons(x, l1), l2)cons(x, app(l1, l2))
app(app(l1, l2), l3)app(l1, app(l2, l3))mem(x, nil)false
mem(x, cons(y, l))ifmem(eq(x, y), x, l)ifmem(true, x, l)true
ifmem(false, x, l)mem(x, l)inter(x, nil)nil
inter(nil, x)nilinter(app(l1, l2), l3)app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3))app(inter(l1, l2), inter(l1, l3))inter(cons(x, l1), l2)ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2))ifinter(mem(x, l1), x, l2, l1)ifinter(true, x, l1, l2)cons(x, inter(l1, l2))
ifinter(false, x, l1, l2)inter(l1, l2)

Original Signature

Termination of terms over the following signature is verified: app, ifmem, true, inter, 0, s, if, mem, false, ifinter, eq, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

app#(cons(x, l1), l2)app#(l1, l2)app#(app(l1, l2), l3)app#(l2, l3)
app#(app(l1, l2), l3)app#(l1, app(l2, l3))

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

ifmem#(false, x, l)mem#(x, l)mem#(x, cons(y, l))ifmem#(eq(x, y), x, l)

Rewrite Rules

if(true, x, y)xif(false, x, y)y
eq(0, 0)trueeq(0, s(x))false
eq(s(x), 0)falseeq(s(x), s(y))eq(x, y)
app(nil, l)lapp(cons(x, l1), l2)cons(x, app(l1, l2))
app(app(l1, l2), l3)app(l1, app(l2, l3))mem(x, nil)false
mem(x, cons(y, l))ifmem(eq(x, y), x, l)ifmem(true, x, l)true
ifmem(false, x, l)mem(x, l)inter(x, nil)nil
inter(nil, x)nilinter(app(l1, l2), l3)app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3))app(inter(l1, l2), inter(l1, l3))inter(cons(x, l1), l2)ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2))ifinter(mem(x, l1), x, l2, l1)ifinter(true, x, l1, l2)cons(x, inter(l1, l2))
ifinter(false, x, l1, l2)inter(l1, l2)

Original Signature

Termination of terms over the following signature is verified: app, ifmem, true, inter, 0, s, if, mem, false, ifinter, eq, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

mem#(x, cons(y, l))ifmem#(eq(x, y), x, l)

Problem 6: DependencyGraph



Dependency Pair Problem

Dependency Pairs

ifmem#(false, x, l)mem#(x, l)

Rewrite Rules

if(true, x, y)xif(false, x, y)y
eq(0, 0)trueeq(0, s(x))false
eq(s(x), 0)falseeq(s(x), s(y))eq(x, y)
app(nil, l)lapp(cons(x, l1), l2)cons(x, app(l1, l2))
app(app(l1, l2), l3)app(l1, app(l2, l3))mem(x, nil)false
mem(x, cons(y, l))ifmem(eq(x, y), x, l)ifmem(true, x, l)true
ifmem(false, x, l)mem(x, l)inter(x, nil)nil
inter(nil, x)nilinter(app(l1, l2), l3)app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3))app(inter(l1, l2), inter(l1, l3))inter(cons(x, l1), l2)ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2))ifinter(mem(x, l1), x, l2, l1)ifinter(true, x, l1, l2)cons(x, inter(l1, l2))
ifinter(false, x, l1, l2)inter(l1, l2)

Original Signature

Termination of terms over the following signature is verified: app, ifmem, true, inter, 0, s, if, mem, false, ifinter, eq, nil, cons

Strategy


There are no SCCs!

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

inter#(l1, app(l2, l3))inter#(l1, l3)inter#(l1, app(l2, l3))inter#(l1, l2)
ifinter#(true, x, l1, l2)inter#(l1, l2)inter#(l1, cons(x, l2))ifinter#(mem(x, l1), x, l2, l1)
inter#(app(l1, l2), l3)inter#(l2, l3)ifinter#(false, x, l1, l2)inter#(l1, l2)
inter#(cons(x, l1), l2)ifinter#(mem(x, l2), x, l1, l2)inter#(app(l1, l2), l3)inter#(l1, l3)

Rewrite Rules

if(true, x, y)xif(false, x, y)y
eq(0, 0)trueeq(0, s(x))false
eq(s(x), 0)falseeq(s(x), s(y))eq(x, y)
app(nil, l)lapp(cons(x, l1), l2)cons(x, app(l1, l2))
app(app(l1, l2), l3)app(l1, app(l2, l3))mem(x, nil)false
mem(x, cons(y, l))ifmem(eq(x, y), x, l)ifmem(true, x, l)true
ifmem(false, x, l)mem(x, l)inter(x, nil)nil
inter(nil, x)nilinter(app(l1, l2), l3)app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3))app(inter(l1, l2), inter(l1, l3))inter(cons(x, l1), l2)ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2))ifinter(mem(x, l1), x, l2, l1)ifinter(true, x, l1, l2)cons(x, inter(l1, l2))
ifinter(false, x, l1, l2)inter(l1, l2)

Original Signature

Termination of terms over the following signature is verified: app, ifmem, true, inter, 0, s, if, mem, false, ifinter, eq, nil, cons

Strategy


Polynomial Interpretation

Improved Usable rules

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

ifinter#(true, x, l1, l2)inter#(l1, l2)inter#(l1, cons(x, l2))ifinter#(mem(x, l1), x, l2, l1)
ifinter#(false, x, l1, l2)inter#(l1, l2)inter#(cons(x, l1), l2)ifinter#(mem(x, l2), x, l1, l2)

Problem 7: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

inter#(l1, app(l2, l3))inter#(l1, l3)inter#(l1, app(l2, l3))inter#(l1, l2)
inter#(app(l1, l2), l3)inter#(l2, l3)inter#(app(l1, l2), l3)inter#(l1, l3)

Rewrite Rules

if(true, x, y)xif(false, x, y)y
eq(0, 0)trueeq(0, s(x))false
eq(s(x), 0)falseeq(s(x), s(y))eq(x, y)
app(nil, l)lapp(cons(x, l1), l2)cons(x, app(l1, l2))
app(app(l1, l2), l3)app(l1, app(l2, l3))mem(x, nil)false
mem(x, cons(y, l))ifmem(eq(x, y), x, l)ifmem(true, x, l)true
ifmem(false, x, l)mem(x, l)inter(x, nil)nil
inter(nil, x)nilinter(app(l1, l2), l3)app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3))app(inter(l1, l2), inter(l1, l3))inter(cons(x, l1), l2)ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2))ifinter(mem(x, l1), x, l2, l1)ifinter(true, x, l1, l2)cons(x, inter(l1, l2))
ifinter(false, x, l1, l2)inter(l1, l2)

Original Signature

Termination of terms over the following signature is verified: app, ifmem, true, inter, 0, s, if, mem, false, ifinter, eq, nil, cons

Strategy


Polynomial Interpretation

There are no usable rules

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

inter#(app(l1, l2), l3)inter#(l2, l3)inter#(app(l1, l2), l3)inter#(l1, l3)

Problem 8: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

inter#(l1, app(l2, l3))inter#(l1, l3)inter#(l1, app(l2, l3))inter#(l1, l2)

Rewrite Rules

if(true, x, y)xif(false, x, y)y
eq(0, 0)trueeq(0, s(x))false
eq(s(x), 0)falseeq(s(x), s(y))eq(x, y)
app(nil, l)lapp(cons(x, l1), l2)cons(x, app(l1, l2))
app(app(l1, l2), l3)app(l1, app(l2, l3))mem(x, nil)false
mem(x, cons(y, l))ifmem(eq(x, y), x, l)ifmem(true, x, l)true
ifmem(false, x, l)mem(x, l)inter(x, nil)nil
inter(nil, x)nilinter(app(l1, l2), l3)app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3))app(inter(l1, l2), inter(l1, l3))inter(cons(x, l1), l2)ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2))ifinter(mem(x, l1), x, l2, l1)ifinter(true, x, l1, l2)cons(x, inter(l1, l2))
ifinter(false, x, l1, l2)inter(l1, l2)

Original Signature

Termination of terms over the following signature is verified: app, ifmem, true, inter, 0, s, if, mem, false, ifinter, eq, nil, cons

Strategy


Polynomial Interpretation

There are no usable rules

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

inter#(l1, app(l2, l3))inter#(l1, l3)inter#(l1, app(l2, l3))inter#(l1, l2)