TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60000 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (64ms).
 | – Problem 2 was processed with processor BackwardInstantiation (2ms).
 |    | – Problem 5 was processed with processor BackwardInstantiation (4ms).
 |    |    | – Problem 6 was processed with processor Propagation (2ms).
 |    |    |    | – Problem 7 remains open; application of the following processors failed [ForwardNarrowing (0ms), BackwardInstantiation (1ms), ForwardInstantiation (2ms), Propagation (0ms)].
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

cond#(true, x, y)minus#(x, s(y))minus#(x, y)cond#(equal(min(x, y), y), x, y)

Rewrite Rules

minus(x, y)cond(equal(min(x, y), y), x, y)cond(true, x, y)s(minus(x, s(y)))
min(0, v)0min(u, 0)0
min(s(u), s(v))s(min(u, v))equal(0, 0)true
equal(s(x), 0)falseequal(0, s(y))false
equal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: min, 0, minus, s, false, true, equal, cond


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

minus#(x, y)min#(x, y)cond#(true, x, y)minus#(x, s(y))
minus#(x, y)equal#(min(x, y), y)equal#(s(x), s(y))equal#(x, y)
minus#(x, y)cond#(equal(min(x, y), y), x, y)min#(s(u), s(v))min#(u, v)

Rewrite Rules

minus(x, y)cond(equal(min(x, y), y), x, y)cond(true, x, y)s(minus(x, s(y)))
min(0, v)0min(u, 0)0
min(s(u), s(v))s(min(u, v))equal(0, 0)true
equal(s(x), 0)falseequal(0, s(y))false
equal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: min, minus, 0, s, true, false, equal, cond

Strategy


The following SCCs where found

equal#(s(x), s(y)) → equal#(x, y)

cond#(true, x, y) → minus#(x, s(y))minus#(x, y) → cond#(equal(min(x, y), y), x, y)

min#(s(u), s(v)) → min#(u, v)

Problem 2: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

cond#(true, x, y)minus#(x, s(y))minus#(x, y)cond#(equal(min(x, y), y), x, y)

Rewrite Rules

minus(x, y)cond(equal(min(x, y), y), x, y)cond(true, x, y)s(minus(x, s(y)))
min(0, v)0min(u, 0)0
min(s(u), s(v))s(min(u, v))equal(0, 0)true
equal(s(x), 0)falseequal(0, s(y))false
equal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: min, minus, 0, s, true, false, equal, cond

Strategy


Instantiation

For all potential predecessors l → r of the rule minus#(x, y) → cond#(equal(min(x, y), y), x, y) on dependency pair chains it holds that: Thus, minus#(x, y) → cond#(equal(min(x, y), y), x, y) is replaced by instances determined through the above matching. These instances are:
minus#(_x, s(_y)) → cond#(equal(min(_x, s(_y)), s(_y)), _x, s(_y))

Problem 5: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

cond#(true, x, y)minus#(x, s(y))minus#(_x, s(_y))cond#(equal(min(_x, s(_y)), s(_y)), _x, s(_y))

Rewrite Rules

minus(x, y)cond(equal(min(x, y), y), x, y)cond(true, x, y)s(minus(x, s(y)))
min(0, v)0min(u, 0)0
min(s(u), s(v))s(min(u, v))equal(0, 0)true
equal(s(x), 0)falseequal(0, s(y))false
equal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: min, 0, minus, s, false, true, equal, cond

Strategy


Instantiation

For all potential predecessors l → r of the rule minus#(_x, s(_y)) → cond#(equal(min(_x, s(_y)), s(_y)), _x, s(_y)) on dependency pair chains it holds that: Thus, minus#(_x, s(_y)) → cond#(equal(min(_x, s(_y)), s(_y)), _x, s(_y)) is replaced by instances determined through the above matching. These instances are:
minus#(x, s(y)) → cond#(equal(min(x, s(y)), s(y)), x, s(y))

Problem 6: Propagation



Dependency Pair Problem

Dependency Pairs

cond#(true, x, y)minus#(x, s(y))minus#(x, s(y))cond#(equal(min(x, s(y)), s(y)), x, s(y))

Rewrite Rules

minus(x, y)cond(equal(min(x, y), y), x, y)cond(true, x, y)s(minus(x, s(y)))
min(0, v)0min(u, 0)0
min(s(u), s(v))s(min(u, v))equal(0, 0)true
equal(s(x), 0)falseequal(0, s(y))false
equal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: min, minus, 0, s, true, false, equal, cond

Strategy


The dependency pairs cond#(true, x, y) → minus#(x, s(y)) and minus#(x, s(y)) → cond#(equal(min(x, s(y)), s(y)), x, s(y)) are consolidated into the rule cond#(true, x, y) → cond#(equal(min(x, s(y)), s(y)), x, s(y)) .

This is possible as

The dependency pairs cond#(true, x, y) → minus#(x, s(y)) and minus#(x, s(y)) → cond#(equal(min(x, s(y)), s(y)), x, s(y)) are consolidated into the rule cond#(true, x, y) → cond#(equal(min(x, s(y)), s(y)), x, s(y)) .

This is possible as


Summary

Removed Dependency PairsAdded Dependency Pairs
cond#(true, x, y) → minus#(x, s(y))cond#(true, x, y) → cond#(equal(min(x, s(y)), s(y)), x, s(y))
minus#(x, s(y)) → cond#(equal(min(x, s(y)), s(y)), x, s(y)) 

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

min#(s(u), s(v))min#(u, v)

Rewrite Rules

minus(x, y)cond(equal(min(x, y), y), x, y)cond(true, x, y)s(minus(x, s(y)))
min(0, v)0min(u, 0)0
min(s(u), s(v))s(min(u, v))equal(0, 0)true
equal(s(x), 0)falseequal(0, s(y))false
equal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: min, minus, 0, s, true, false, equal, cond

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

min#(s(u), s(v))min#(u, v)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

equal#(s(x), s(y))equal#(x, y)

Rewrite Rules

minus(x, y)cond(equal(min(x, y), y), x, y)cond(true, x, y)s(minus(x, s(y)))
min(0, v)0min(u, 0)0
min(s(u), s(v))s(min(u, v))equal(0, 0)true
equal(s(x), 0)falseequal(0, s(y))false
equal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: min, minus, 0, s, true, false, equal, cond

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

equal#(s(x), s(y))equal#(x, y)