TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60002 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (50ms).
 | – 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 (1ms), DependencyGraph (3ms), PolynomialLinearRange4iUR (330ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (2676ms), DependencyGraph (5ms), ReductionPairSAT (timeout)].

The following open problems remain:



Open Dependency Pair Problem 4

Dependency Pairs

if#(true, x, l, accu, orig)rev#(s(x), tail(l), cons(head(l), accu), orig)rev#(x, l, accu, orig)if#(lt(x, length(orig)), x, l, accu, orig)

Rewrite Rules

length(nil)0length(cons(x, l))s(length(l))
lt(x, 0)falselt(0, s(y))true
lt(s(x), s(y))lt(x, y)head(cons(x, l))x
head(nil)undefinedtail(nil)nil
tail(cons(x, l))lreverse(l)rev(0, l, nil, l)
rev(x, l, accu, orig)if(lt(x, length(orig)), x, l, accu, orig)if(true, x, l, accu, orig)rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig)accu

Original Signature

Termination of terms over the following signature is verified: rev, reverse, true, lt, tail, 0, s, if, undefined, length, false, head, nil, cons


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if#(true, x, l, accu, orig)head#(l)reverse#(l)rev#(0, l, nil, l)
if#(true, x, l, accu, orig)rev#(s(x), tail(l), cons(head(l), accu), orig)rev#(x, l, accu, orig)if#(lt(x, length(orig)), x, l, accu, orig)
rev#(x, l, accu, orig)lt#(x, length(orig))rev#(x, l, accu, orig)length#(orig)
lt#(s(x), s(y))lt#(x, y)length#(cons(x, l))length#(l)
if#(true, x, l, accu, orig)tail#(l)

Rewrite Rules

length(nil)0length(cons(x, l))s(length(l))
lt(x, 0)falselt(0, s(y))true
lt(s(x), s(y))lt(x, y)head(cons(x, l))x
head(nil)undefinedtail(nil)nil
tail(cons(x, l))lreverse(l)rev(0, l, nil, l)
rev(x, l, accu, orig)if(lt(x, length(orig)), x, l, accu, orig)if(true, x, l, accu, orig)rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig)accu

Original Signature

Termination of terms over the following signature is verified: rev, reverse, true, lt, tail, 0, s, if, undefined, length, false, head, cons, nil

Strategy


The following SCCs where found

lt#(s(x), s(y)) → lt#(x, y)

length#(cons(x, l)) → length#(l)

if#(true, x, l, accu, orig) → rev#(s(x), tail(l), cons(head(l), accu), orig)rev#(x, l, accu, orig) → if#(lt(x, length(orig)), x, l, accu, orig)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

lt#(s(x), s(y))lt#(x, y)

Rewrite Rules

length(nil)0length(cons(x, l))s(length(l))
lt(x, 0)falselt(0, s(y))true
lt(s(x), s(y))lt(x, y)head(cons(x, l))x
head(nil)undefinedtail(nil)nil
tail(cons(x, l))lreverse(l)rev(0, l, nil, l)
rev(x, l, accu, orig)if(lt(x, length(orig)), x, l, accu, orig)if(true, x, l, accu, orig)rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig)accu

Original Signature

Termination of terms over the following signature is verified: rev, reverse, true, lt, tail, 0, s, if, undefined, length, false, head, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

lt#(s(x), s(y))lt#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

length#(cons(x, l))length#(l)

Rewrite Rules

length(nil)0length(cons(x, l))s(length(l))
lt(x, 0)falselt(0, s(y))true
lt(s(x), s(y))lt(x, y)head(cons(x, l))x
head(nil)undefinedtail(nil)nil
tail(cons(x, l))lreverse(l)rev(0, l, nil, l)
rev(x, l, accu, orig)if(lt(x, length(orig)), x, l, accu, orig)if(true, x, l, accu, orig)rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig)accu

Original Signature

Termination of terms over the following signature is verified: rev, reverse, true, lt, tail, 0, s, if, undefined, length, false, head, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

length#(cons(x, l))length#(l)