=== Bilinearity of $\oplus'$ === First we need to show that $\oplus'$ is bilinear. $(1): \quad$ Let $\left({a_1, b_1}\right), \left({a_2, b_2}\right), \left({c, d}\right) \in A'$. Then: {{begin-eqn}} {{eqn | o =        | r = \left({\left({a_1, b_1}\right) + \left({a_2, b_2}\right)}\right) \oplus' \left({c, d}\right)       | c =  }} {{eqn | r = \left({a_1 + a_2, b_1 + b_2}\right) \oplus' \left({c, d}\right)       | c =  }} {{eqn | r = \left({\left({a_1 + a_2}\right) \oplus c - d \oplus \left({b_1 + b_2}\right)^*, \left({a_1 + a_2}\right)^* \oplus d + c \oplus \left({b_1 + b_2}\right)}\right)       | c =  }} {{eqn | r = \left({a_1 \oplus c + a_2 \oplus c - d \oplus b_1^* - d \oplus b_2^*, a_1^* \oplus d + a_2^* \oplus d + c \oplus b_1 + c \oplus b_2}\right)       | c = as $\oplus$ is bilinear }} {{eqn | r = \left({a_1 \oplus c - d \oplus b_1^*, a_1^* \oplus d + c \oplus b_1}\right) + \left({a_2 \oplus c - d \oplus b_2^*, a_2^* \oplus d + c \oplus b_2}\right)       | c =  }} {{eqn | r = \left({\left({a_1, b_1}\right) \oplus' \left({c, d}\right)}\right) + \left({\left({a_2, b_2}\right) \oplus' \left({c, d}\right)}\right)       | c =  }} {{end-eqn}} Similarly (and equally tediously) we can show that: :$\left({c, d}\right) \oplus' \left({\left({a_1, b_1}\right) + \left({a_2, b_2}\right)}\right) = \left({\left({c, d}\right) \oplus' \left({a_1, b_1}\right)}\right) + \left({\left({c, d}\right) \oplus' \left({a_2, b_2}\right)}\right)$ $(2): \quad$ Let $\left({a, b}\right), \left({c, d}\right) \in A'$ and $\alpha, \beta \in \R$. Then: {{begin-eqn}} {{eqn | o =        | r = \left({\alpha \left({a, b}\right)}\right) \oplus' \left({c, d}\right)       | c =  }} {{eqn | r = \left({\alpha a, \alpha b}\right) \oplus' \left({c, d}\right)       | c = as $\oplus$ is bilinear }} {{eqn | r = \left({\alpha a \oplus c - d \oplus \alpha b^*, \alpha a^* \oplus d + c \oplus \alpha b}\right)       | c =  }} {{eqn | r = \alpha \left({a \oplus c - d \oplus b^*, a^* \oplus d + c \oplus b}\right)       | c = as $\oplus$ is bilinear }} {{eqn | r = \alpha \left({a, b}\right) \oplus' \left({c, d}\right)       | c =  }} {{end-eqn}} Similarly: :$\left({a, b}\right) \oplus' \left({\left({c, d}\right) \beta}\right) = \left({a, b}\right) \oplus' \left({c, d}\right) \beta$ So $\oplus'$ has been shown to be a bilinear mapping. === Conjugate Nature of $*'$ === We have that: :$\forall \left({a, b}\right) \in A': {\left({a, b}\right)^*}' = \left({a^*, -b}\right)$ So: {{begin-eqn}} {{eqn | o  =       | r = {\left({ {\left({a, b}\right)^*}'}\right)^*}'       | c =  }} {{eqn | r = {\left({a^*, -b}\right)^*}'       | c =  }} {{eqn | r = \left({\left({a^*}\right)^*, -\left({-b}\right)}\right)       | c =  }} {{eqn | r = \left({a, b}\right)       | c = as $*$ is a conjugation on $A$ }} {{end-eqn}} Finally: {{begin-eqn}} {{eqn | o =        | r = {\left({\left({a, b}\right) \oplus' \left({c, d}\right)}\right)^*}'       | c =  }} {{eqn | r = {\left({a \oplus c - d \oplus b^*, a^* \oplus d + c \oplus b}\right)^*}'       | c =  }} {{eqn | r = \left({\left({a \oplus c - d \oplus b^*}\right)^*, -\left({a^* \oplus d + c \oplus b}\right)}\right)       | c =  }} {{eqn | r = \left({c^* \oplus a^* - b \oplus d^*, -\left({c \oplus b + a^* \oplus d}\right)}\right)       | c =  }} {{eqn | r = \left({c^*, -d}\right) \oplus' \left({a^*, -b}\right)       | c =  }} {{eqn | r = {\left({c, d}\right)^*}' \oplus' {\left({a, b}\right)^*}'       | c =  }} {{end-eqn}} thus proving that $*'$ is a conjugation on $A'$. Hence the result. {{qed}}