The rewrite relation of the following TRS is considered.
sel(s(X),cons(Y,Z)) |
→ |
sel(X,activate(Z)) |
(1) |
sel(0,cons(X,Z)) |
→ |
X |
(2) |
first(0,Z) |
→ |
nil |
(3) |
first(s(X),cons(Y,Z)) |
→ |
cons(Y,n__first(X,activate(Z))) |
(4) |
from(X) |
→ |
cons(X,n__from(s(X))) |
(5) |
sel1(s(X),cons(Y,Z)) |
→ |
sel1(X,activate(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,activate(Z))) |
(9) |
quote(n__0) |
→ |
01 |
(10) |
quote1(n__cons(X,Z)) |
→ |
cons1(quote(activate(X)),quote1(activate(Z))) |
(11) |
quote1(n__nil) |
→ |
nil1 |
(12) |
quote(n__s(X)) |
→ |
s1(quote(activate(X))) |
(13) |
quote(n__sel(X,Z)) |
→ |
sel1(activate(X),activate(Z)) |
(14) |
quote1(n__first(X,Z)) |
→ |
first1(activate(X),activate(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) |
first(X1,X2) |
→ |
n__first(X1,X2) |
(21) |
from(X) |
→ |
n__from(X) |
(22) |
0 |
→ |
n__0 |
(23) |
cons(X1,X2) |
→ |
n__cons(X1,X2) |
(24) |
nil |
→ |
n__nil |
(25) |
s(X) |
→ |
n__s(X) |
(26) |
sel(X1,X2) |
→ |
n__sel(X1,X2) |
(27) |
activate(n__first(X1,X2)) |
→ |
first(X1,X2) |
(28) |
activate(n__from(X)) |
→ |
from(X) |
(29) |
activate(n__0) |
→ |
0 |
(30) |
activate(n__cons(X1,X2)) |
→ |
cons(X1,X2) |
(31) |
activate(n__nil) |
→ |
nil |
(32) |
activate(n__s(X)) |
→ |
s(X) |
(33) |
activate(n__sel(X1,X2)) |
→ |
sel(X1,X2) |
(34) |
activate(X) |
→ |
X |
(35) |
The evaluation strategy is outermost