Bridgeless cubic graph has a 1-factor not containing two arbitrarily prescribed lines According to Petersen's theorem, every bridgeless cubic graph has a perfect matching.
While studying the proof of Petersen's theorem I came accross the following theorem
"every bridgeless cubic graph has a 1-factor not containing two arbitrarily prescribed lines", which I found in the PDF attached herewith (page 2, paragraph 3). Although they have mentioned the author who has proved it, the article is not in English but I would like to study the proof of this theorem as well.
Can someone help me in this regard? How to prove the theorem "every bridgeless cubic graph has a 1-factor not containing two arbitrarily prescribed lines"?
Thanks a lot in advance.
https://dml.cz/bitstream/handle/10338.dmlcz/127027/MathSlov_22-1972-4_8.pdf
 A: Let the bridgeless cubic graph be $G=(V,E)$ and $G'=(V,E')$ be the graph where two arbitrary edges are removed.
By Tutte's Theorem, if $G'$ has no perfect matchings, we have some set $U\subset V$ such that the odd components $O_i,\cdots,O_n$ of $G'-U$ satisfying $n > |U|$. Further, $n$ and $|U|$ must be of the same parity since $|V|$ is even. Therefore $n\geq |U|+2$.
Denote the edges (in $G$) connecting $V$ to $O_i$ be $s_i$ and the edges connecting $O_i$ to the other odd components be $t_i$. 
Since only 2 edges are removed, and $t_i$'s can only arise from these edges, we have $\sum_{i=1}^n t_i \leq 4$.
Next, since the graph is 3-regular, the sum over all $s_i$'s must not be greater than $3|U|$ and we have $\sum_{i=1}^n s_i \leq 3|U|$.
Now, note that the graph is bridgeless. Hence there must be more than one edge connecting each component $V_i$ to other components. The handshaking lemma necessitates the edges inside the component $V_i$ to be even (since it is also a graph) and the sum of all degrees of $V_i$ in $G$ is odd. Therefore, the edges not within the component will be at least 3. (i.e. $s_i+t_i\geq 3$)
Combining all 3 inequalities, we get $3n=3\sum_{i=1}^n s_i+t_i\leq 4+3|U|$. Rearranging, we get $3(|U|+2-n)-2\geq 0$.
This is a contradiction to Tutte's theorem since $n\geq |U|+2$. Therefore, $G'$ must have a perfect matching.
