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)
abac

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)
abac

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)
abac

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)
abac

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)
abac

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)