The rewrite relation of the following TRS is considered.
sel(s(X),cons(Y,Z)) |
→ |
sel(X,Z) |
(1) |
sel(0,cons(X,Z)) |
→ |
X |
(2) |
first(0,Z) |
→ |
nil |
(3) |
first(s(X),cons(Y,Z)) |
→ |
cons(Y,first(X,Z)) |
(4) |
from(X) |
→ |
cons(X,from(s(X))) |
(5) |
sel1(s(X),cons(Y,Z)) |
→ |
sel1(X,Z) |
(6) |
sel1(0,cons(X,Z)) |
→ |
quote(X) |
(7) |
first1(0,Z) |
→ |
nil1 |
(8) |
first1(s(X),cons(Y,Z)) |
→ |
cons1(quote(Y),first1(X,Z)) |
(9) |
quote(0) |
→ |
01 |
(10) |
quote1(cons(X,Z)) |
→ |
cons1(quote(X),quote1(Z)) |
(11) |
quote1(nil) |
→ |
nil1 |
(12) |
quote(s(X)) |
→ |
s1(quote(X)) |
(13) |
quote(sel(X,Z)) |
→ |
sel1(X,Z) |
(14) |
quote1(first(X,Z)) |
→ |
first1(X,Z) |
(15) |
unquote(01) |
→ |
0 |
(16) |
unquote(s1(X)) |
→ |
s(unquote(X)) |
(17) |
unquote1(nil1) |
→ |
nil |
(18) |
unquote1(cons1(X,Z)) |
→ |
fcons(unquote(X),unquote1(Z)) |
(19) |
fcons(X,Z) |
→ |
cons(X,Z) |
(20) |
The evaluation strategy is outermost