YES
  TPA v.1.0

Result: TRS is terminating

Default interpretations for symbols are not printed. For polynomial interpretations
and semantic labelling over N\{0,1} defaults are 2 for constants, identity for unary
symbols and x+y-2 for binary symbols. For semantic labelling over {0,1} (booleans)
defaults are 0 for constants, identity for unary symbols and disjunction for binary symbols.

[1]  TRS loaded from input file:
(1) top(ok(new(x))) -> top(check(x))
(2) top(ok(old(x))) -> top(check(x))
(3) bot ->= new(bot)
(4) check(new(x)) ->= new(check(x))
(5) check(old(x)) ->= old(check(x))
(6) check(old(x)) ->= ok(old(x))
(7) new(ok(x)) ->= ok(new(x))
(8) old(ok(x)) ->= ok(old(x))

[2]  Use following polynomial interpretation:
[old(x)] = x + 1
rest default

  Remove rules with left hand side strictly bigger than right hand side:
(2)

[3]  Label this TRS using following interpretation that is a model:
[top(x)] = 0
[old(x)] = 1
rest default

  thus obtaining new TRS:
(1a) top$1(ok$1(new$1(x))) -> top$1(check$1(x))
(1b) top$0(ok$0(new$0(x))) -> top$0(check$0(x))
(3a) bot ->= new$0(bot)
(4a) check$1(new$1(x)) ->= new$1(check$1(x))
(4b) check$0(new$0(x)) ->= new$0(check$0(x))
(5a) check$1(old$1(x)) ->= old$1(check$1(x))
(5b) check$1(old$0(x)) ->= old$0(check$0(x))
(6a) check$1(old$1(x)) ->= ok$1(old$1(x))
(6b) check$1(old$0(x)) ->= ok$1(old$0(x))
(7a) new$1(ok$1(x)) ->= ok$1(new$1(x))
(7b) new$0(ok$0(x)) ->= ok$0(new$0(x))
(8a) old$1(ok$1(x)) ->= ok$1(old$1(x))
(8b) old$0(ok$0(x)) ->= ok$1(old$0(x))

[4]  Use following polynomial interpretation:
[ok$0(x)] = x + 1
rest default

  Remove rules with left hand side strictly bigger than right hand side:
(1b), (8b)

[5]  Use following polynomial interpretation:
[new$1(x)] = x + 1
rest default

  Remove rules with left hand side strictly bigger than right hand side:
(1a)

[6]  Since there are no remaining strict rules, relative termination is proved!

../tpdb/TRS/relative/rtL-wl1nz.trs, 0.01, Y
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