TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60020 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (45ms).
| – Problem 2 was processed with processor BackwardInstantiation (2ms).
| | – Problem 5 remains open; application of the following processors failed [ForwardInstantiation (4ms), Propagation (1ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (3ms), Propagation (1ms)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| help#(x, y) | → | ifb#(lt(y, x), x, y) | | ifb#(true, x, y) | → | help#(half(x), s(y)) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | logarithm(x) | → | ifa(lt(0, x), x) |
| ifa(true, x) | → | help(x, 1) | | ifa(false, x) | → | logZeroError |
| help(x, y) | → | ifb(lt(y, x), x, y) | | ifb(true, x, y) | → | help(half(x), s(y)) |
| ifb(false, x, y) | → | y | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(x))) | → | s(half(x)) |
Original Signature
Termination of terms over the following signature is verified: 1, help, 0, s, logarithm, half, false, true, lt, logZeroError, ifb, ifa
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| logarithm#(x) | → | lt#(0, x) | | logarithm#(x) | → | ifa#(lt(0, x), x) |
| ifa#(true, x) | → | help#(x, 1) | | half#(s(s(x))) | → | half#(x) |
| ifb#(true, x, y) | → | half#(x) | | lt#(s(x), s(y)) | → | lt#(x, y) |
| help#(x, y) | → | ifb#(lt(y, x), x, y) | | help#(x, y) | → | lt#(y, x) |
| ifb#(true, x, y) | → | help#(half(x), s(y)) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | logarithm(x) | → | ifa(lt(0, x), x) |
| ifa(true, x) | → | help(x, 1) | | ifa(false, x) | → | logZeroError |
| help(x, y) | → | ifb(lt(y, x), x, y) | | ifb(true, x, y) | → | help(half(x), s(y)) |
| ifb(false, x, y) | → | y | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(x))) | → | s(half(x)) |
Original Signature
Termination of terms over the following signature is verified: help, 1, 0, s, logarithm, true, false, half, lt, logZeroError, ifb, ifa
Strategy
The following SCCs where found
| help#(x, y) → ifb#(lt(y, x), x, y) | ifb#(true, x, y) → help#(half(x), s(y)) |
| half#(s(s(x))) → half#(x) |
| lt#(s(x), s(y)) → lt#(x, y) |
Problem 2: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
| help#(x, y) | → | ifb#(lt(y, x), x, y) | | ifb#(true, x, y) | → | help#(half(x), s(y)) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | logarithm(x) | → | ifa(lt(0, x), x) |
| ifa(true, x) | → | help(x, 1) | | ifa(false, x) | → | logZeroError |
| help(x, y) | → | ifb(lt(y, x), x, y) | | ifb(true, x, y) | → | help(half(x), s(y)) |
| ifb(false, x, y) | → | y | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(x))) | → | s(half(x)) |
Original Signature
Termination of terms over the following signature is verified: help, 1, 0, s, logarithm, true, false, half, lt, logZeroError, ifb, ifa
Strategy
Instantiation
For all potential predecessors l → r of the rule help
#(
x,
y) → ifb
#(lt(
y,
x),
x,
y) on dependency pair chains it holds that:
- help#(x, y) matches r,
- all variables of help#(x, y) are embedded in constructor contexts, i.e., each subterm of help#(x, y), containing a variable is rooted by a constructor symbol.
Thus, help
#(
x,
y) → ifb
#(lt(
y,
x),
x,
y) is replaced by instances determined through the above matching. These instances are:
| help#(half(_x), s(_y)) → ifb#(lt(s(_y), half(_x)), half(_x), s(_y)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| half#(s(s(x))) | → | half#(x) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | logarithm(x) | → | ifa(lt(0, x), x) |
| ifa(true, x) | → | help(x, 1) | | ifa(false, x) | → | logZeroError |
| help(x, y) | → | ifb(lt(y, x), x, y) | | ifb(true, x, y) | → | help(half(x), s(y)) |
| ifb(false, x, y) | → | y | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(x))) | → | s(half(x)) |
Original Signature
Termination of terms over the following signature is verified: help, 1, 0, s, logarithm, true, false, half, lt, logZeroError, ifb, ifa
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| half#(s(s(x))) | → | half#(x) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| lt#(s(x), s(y)) | → | lt#(x, y) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | logarithm(x) | → | ifa(lt(0, x), x) |
| ifa(true, x) | → | help(x, 1) | | ifa(false, x) | → | logZeroError |
| help(x, y) | → | ifb(lt(y, x), x, y) | | ifb(true, x, y) | → | help(half(x), s(y)) |
| ifb(false, x, y) | → | y | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(x))) | → | s(half(x)) |
Original Signature
Termination of terms over the following signature is verified: help, 1, 0, s, logarithm, true, false, half, lt, logZeroError, ifb, ifa
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| lt#(s(x), s(y)) | → | lt#(x, y) |