MAYBE

Problem:
 .(1(),x) -> x
 .(x,1()) -> x
 .(i(x),x) -> 1()
 .(x,i(x)) -> 1()
 i(1()) -> 1()
 i(i(x)) -> x
 .(i(y),.(y,z)) -> z
 .(y,.(i(y),z)) -> z
 .(.(x,y),z) -> .(x,.(y,z))
 i(.(x,y)) -> .(i(y),i(x))

Proof:
 Open