The rewrite relation of the following TRS is considered.
app(app(.,1),x) |
→ |
x |
(1) |
app(app(.,x),1) |
→ |
x |
(2) |
app(app(.,app(i,x)),x) |
→ |
1 |
(3) |
app(app(.,x),app(i,x)) |
→ |
1 |
(4) |
app(app(.,app(i,y)),app(app(.,y),z)) |
→ |
z |
(5) |
app(app(.,y),app(app(.,app(i,y)),z)) |
→ |
z |
(6) |
app(i,1) |
→ |
1 |
(7) |
app(i,app(i,x)) |
→ |
x |
(8) |
app(app(map,f),nil) |
→ |
nil |
(9) |
app(app(map,f),app(app(cons,x),xs)) |
→ |
app(app(cons,app(f,x)),app(app(map,f),xs)) |
(10) |
app(app(filter,f),nil) |
→ |
nil |
(11) |
app(app(filter,f),app(app(cons,x),xs)) |
→ |
app(app(app(app(filter2,app(f,x)),f),x),xs) |
(12) |
app(app(app(app(filter2,true),f),x),xs) |
→ |
app(app(cons,x),app(app(filter,f),xs)) |
(13) |
app(app(app(app(filter2,false),f),x),xs) |
→ |
app(app(filter,f),xs) |
(14) |
app#(app(app(app(filter2,true),f),x),xs) |
→ |
app#(cons,x) |
(15) |
app#(app(filter,f),app(app(cons,x),xs)) |
→ |
app#(app(app(filter2,app(f,x)),f),x) |
(16) |
app#(app(map,f),app(app(cons,x),xs)) |
→ |
app#(cons,app(f,x)) |
(17) |
app#(app(filter,f),app(app(cons,x),xs)) |
→ |
app#(f,x) |
(18) |
app#(app(app(app(filter2,false),f),x),xs) |
→ |
app#(filter,f) |
(19) |
app#(app(map,f),app(app(cons,x),xs)) |
→ |
app#(app(cons,app(f,x)),app(app(map,f),xs)) |
(20) |
app#(app(map,f),app(app(cons,x),xs)) |
→ |
app#(f,x) |
(21) |
app#(app(filter,f),app(app(cons,x),xs)) |
→ |
app#(app(filter2,app(f,x)),f) |
(22) |
app#(app(app(app(filter2,false),f),x),xs) |
→ |
app#(app(filter,f),xs) |
(23) |
app#(app(filter,f),app(app(cons,x),xs)) |
→ |
app#(app(app(app(filter2,app(f,x)),f),x),xs) |
(24) |
app#(app(app(app(filter2,true),f),x),xs) |
→ |
app#(app(filter,f),xs) |
(25) |
app#(app(app(app(filter2,true),f),x),xs) |
→ |
app#(app(cons,x),app(app(filter,f),xs)) |
(26) |
app#(app(filter,f),app(app(cons,x),xs)) |
→ |
app#(filter2,app(f,x)) |
(27) |
app#(app(map,f),app(app(cons,x),xs)) |
→ |
app#(app(map,f),xs) |
(28) |
app#(app(app(app(filter2,true),f),x),xs) |
→ |
app#(filter,f) |
(29) |
The dependency pairs are split into 1
component.