TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60017 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (177ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 remains open; application of the following processors failed [SubtermCriterion (2ms), DependencyGraph (18ms), PolynomialLinearRange4iUR (913ms), DependencyGraph (16ms), PolynomialLinearRange8NegiUR (10997ms), DependencyGraph (14ms), ReductionPairSAT (timeout)].
 | – Problem 5 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 4

Dependency Pairs

if1#(false, b1, b2, b3, x, y, i) → if2#(b1, b2, b3, x, y, i)if4#(false, x, y, i) → gcd2#(x, minus(y, x), i)
if3#(false, b3, x, y, i) → gcd2#(minus(x, y), y, i)gcd2#(x, y, i) → if1#(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
if3#(true, b3, x, y, i) → if4#(b3, x, y, i)if2#(false, b2, b3, x, y, i) → if3#(b2, b3, x, y, i)

Rewrite Rules

gcd(x, y) → gcd2(x, y, 0)gcd2(x, y, i) → if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
if1(true, b1, b2, b3, x, y, i) → pair(result(y), neededIterations(i))if1(false, b1, b2, b3, x, y, i) → if2(b1, b2, b3, x, y, i)
if2(true, b2, b3, x, y, i) → pair(result(x), neededIterations(i))if2(false, b2, b3, x, y, i) → if3(b2, b3, x, y, i)
if3(false, b3, x, y, i) → gcd2(minus(x, y), y, i)if3(true, b3, x, y, i) → if4(b3, x, y, i)
if4(false, x, y, i) → gcd2(x, minus(y, x), i)if4(true, x, y, i) → pair(result(x), neededIterations(i))
inc(0) → 0inc(s(i)) → s(inc(i))
le(s(x), 0) → falsele(0, y) → true
le(s(x), s(y)) → le(x, y)minus(x, 0) → x
minus(0, y) → 0minus(s(x), s(y)) → minus(x, y)
a → ba → c

Original Signature

Termination of terms over the following signature is verified: result, minus, b, neededIterations, c, pair, a, if3, if4, true, if1, if2, 0, s, le, inc, false, gcd2, gcd


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if3#(false, b3, x, y, i) → minus#(x, y)inc#(s(i)) → inc#(i)
gcd2#(x, y, i) → inc#(i)gcd#(x, y) → gcd2#(x, y, 0)
gcd2#(x, y, i) → if1#(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))gcd2#(x, y, i) → le#(x, y)
gcd2#(x, y, i) → le#(x, 0)if4#(false, x, y, i) → minus#(y, x)
gcd2#(x, y, i) → le#(y, 0)le#(s(x), s(y)) → le#(x, y)
if1#(false, b1, b2, b3, x, y, i) → if2#(b1, b2, b3, x, y, i)if4#(false, x, y, i) → gcd2#(x, minus(y, x), i)
if3#(false, b3, x, y, i) → gcd2#(minus(x, y), y, i)minus#(s(x), s(y)) → minus#(x, y)
if3#(true, b3, x, y, i) → if4#(b3, x, y, i)gcd2#(x, y, i) → le#(y, x)
if2#(false, b2, b3, x, y, i) → if3#(b2, b3, x, y, i)

Rewrite Rules

gcd(x, y) → gcd2(x, y, 0)gcd2(x, y, i) → if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
if1(true, b1, b2, b3, x, y, i) → pair(result(y), neededIterations(i))if1(false, b1, b2, b3, x, y, i) → if2(b1, b2, b3, x, y, i)
if2(true, b2, b3, x, y, i) → pair(result(x), neededIterations(i))if2(false, b2, b3, x, y, i) → if3(b2, b3, x, y, i)
if3(false, b3, x, y, i) → gcd2(minus(x, y), y, i)if3(true, b3, x, y, i) → if4(b3, x, y, i)
if4(false, x, y, i) → gcd2(x, minus(y, x), i)if4(true, x, y, i) → pair(result(x), neededIterations(i))
inc(0) → 0inc(s(i)) → s(inc(i))
le(s(x), 0) → falsele(0, y) → true
le(s(x), s(y)) → le(x, y)minus(x, 0) → x
minus(0, y) → 0minus(s(x), s(y)) → minus(x, y)
a → ba → c

Original Signature

Termination of terms over the following signature is verified: result, minus, b, neededIterations, c, pair, a, if3, if4, true, if1, if2, 0, inc, le, s, false, gcd2, gcd

Strategy


The following SCCs where found

le#(s(x), s(y)) → le#(x, y)

inc#(s(i)) → inc#(i)

minus#(s(x), s(y)) → minus#(x, y)

if1#(false, b1, b2, b3, x, y, i) → if2#(b1, b2, b3, x, y, i)if4#(false, x, y, i) → gcd2#(x, minus(y, x), i)
if3#(false, b3, x, y, i) → gcd2#(minus(x, y), y, i)if3#(true, b3, x, y, i) → if4#(b3, x, y, i)
gcd2#(x, y, i) → if1#(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))if2#(false, b2, b3, x, y, i) → if3#(b2, b3, x, y, i)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

minus#(s(x), s(y)) → minus#(x, y)

Rewrite Rules

gcd(x, y) → gcd2(x, y, 0)gcd2(x, y, i) → if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
if1(true, b1, b2, b3, x, y, i) → pair(result(y), neededIterations(i))if1(false, b1, b2, b3, x, y, i) → if2(b1, b2, b3, x, y, i)
if2(true, b2, b3, x, y, i) → pair(result(x), neededIterations(i))if2(false, b2, b3, x, y, i) → if3(b2, b3, x, y, i)
if3(false, b3, x, y, i) → gcd2(minus(x, y), y, i)if3(true, b3, x, y, i) → if4(b3, x, y, i)
if4(false, x, y, i) → gcd2(x, minus(y, x), i)if4(true, x, y, i) → pair(result(x), neededIterations(i))
inc(0) → 0inc(s(i)) → s(inc(i))
le(s(x), 0) → falsele(0, y) → true
le(s(x), s(y)) → le(x, y)minus(x, 0) → x
minus(0, y) → 0minus(s(x), s(y)) → minus(x, y)
a → ba → c

Original Signature

Termination of terms over the following signature is verified: result, minus, b, neededIterations, c, pair, a, if3, if4, true, if1, if2, 0, inc, le, s, false, gcd2, gcd

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

minus#(s(x), s(y)) → minus#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

inc#(s(i)) → inc#(i)

Rewrite Rules

gcd(x, y) → gcd2(x, y, 0)gcd2(x, y, i) → if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
if1(true, b1, b2, b3, x, y, i) → pair(result(y), neededIterations(i))if1(false, b1, b2, b3, x, y, i) → if2(b1, b2, b3, x, y, i)
if2(true, b2, b3, x, y, i) → pair(result(x), neededIterations(i))if2(false, b2, b3, x, y, i) → if3(b2, b3, x, y, i)
if3(false, b3, x, y, i) → gcd2(minus(x, y), y, i)if3(true, b3, x, y, i) → if4(b3, x, y, i)
if4(false, x, y, i) → gcd2(x, minus(y, x), i)if4(true, x, y, i) → pair(result(x), neededIterations(i))
inc(0) → 0inc(s(i)) → s(inc(i))
le(s(x), 0) → falsele(0, y) → true
le(s(x), s(y)) → le(x, y)minus(x, 0) → x
minus(0, y) → 0minus(s(x), s(y)) → minus(x, y)
a → ba → c

Original Signature

Termination of terms over the following signature is verified: result, minus, b, neededIterations, c, pair, a, if3, if4, true, if1, if2, 0, inc, le, s, false, gcd2, gcd

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

inc#(s(i)) → inc#(i)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

le#(s(x), s(y)) → le#(x, y)

Rewrite Rules

gcd(x, y) → gcd2(x, y, 0)gcd2(x, y, i) → if1(le(x, 0), le(y, 0), le(x, y), le(y, x), x, y, inc(i))
if1(true, b1, b2, b3, x, y, i) → pair(result(y), neededIterations(i))if1(false, b1, b2, b3, x, y, i) → if2(b1, b2, b3, x, y, i)
if2(true, b2, b3, x, y, i) → pair(result(x), neededIterations(i))if2(false, b2, b3, x, y, i) → if3(b2, b3, x, y, i)
if3(false, b3, x, y, i) → gcd2(minus(x, y), y, i)if3(true, b3, x, y, i) → if4(b3, x, y, i)
if4(false, x, y, i) → gcd2(x, minus(y, x), i)if4(true, x, y, i) → pair(result(x), neededIterations(i))
inc(0) → 0inc(s(i)) → s(inc(i))
le(s(x), 0) → falsele(0, y) → true
le(s(x), s(y)) → le(x, y)minus(x, 0) → x
minus(0, y) → 0minus(s(x), s(y)) → minus(x, y)
a → ba → c

Original Signature

Termination of terms over the following signature is verified: result, minus, b, neededIterations, c, pair, a, if3, if4, true, if1, if2, 0, inc, le, s, false, gcd2, gcd

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

le#(s(x), s(y)) → le#(x, y)