I'm studying for my exams and I'm stuck at Abstract Algebra. We have $R = ( A , B , G ) , S = ( B , C , H ) , R ′ = ( A , B , G ′ )$ as binary relations: $\operatorname{codom} ( R ) = \operatorname{codom} ( R ′ ) = \operatorname{dom} ( S ) $.
And now we have these properties :
$$S \cdot (R \cap R' ) \subseteq (S \cdot R) \cap (S \cdot R' )$$
$$S \cdot (R \cup R' ) = (S \cdot R) \cup (S \cdot R' )$$
I don't understand why the first one is an inclusion and the second one is an equality.
$$S \cdot (R \cap R') = ( A , C , H \cdot (G \cap G') )$$
$$(S \cdot R) \cap (S \cdot R') = ( A , C , H \cdot G \cap H \cdot G' )$$
$$H \cdot G = \{(a, c) \in A \times C \mid (\exists)b \in B : (a, b) \in G \land (b, c) \in H\}$$
$$H · G' = \{(a, c) \in A \times C \mid (\exists)b \in B : (a, b) \in G' \land (b, c) \in H\}$$
$$H \cdot (G \cap G') = \{(a, c) \in A \times C \mid (\exists)b \in B : (a, b) \in G \land (a, b) \in G' \land (b, c) \in H\}$$
$$H \cdot G \cap H \cdot G' = \{(a, c) \in A \times C \mid (\exists)b \in B : (a, b) \in G \land (b, c) \in H\} \cap \{(a, c) \in A \times C \mid (\exists)b \in B : (a, b) \in G' \land (b, c) \in H\}$$
This is same as :
$$H \cdot G \cap H \cdot G' = \{(a, c) \in A \times C \mid (\exists)b \in B : (a, b) \in G \land (b, c) \in H \land (a, b) \in G' \land (b, c) \in H\}$$
And then since $\land$ is commutative and $(b, c) \in H \land (b, c) \in H = (b, c) \in H $:
$$H \cdot G \cap H \cdot G' = \{(a, c) \in A \times C \mid (\exists)b \in B : (a, b) \in G \land (a, b) \in G' \land (b, c) \in H\}$$
$$H \cdot G \cap H \cdot G' = H \cdot (G \cap G')$$
Same goes for "$\cup$" , just change "$\land$" with "$\lor$" .
What am I missing
PS : If you are downvoting this question ,atleast tell me why.
Here is a counterexample to equality in the first case; you'll see by inspecting it where the problem lies in your argument -- in short " you can't pick the same $b$. "
$$ A:=\{0\}; B:=\{1,1'\}; C:=\{2\}\\ G=\{(0,1)\}; G':=\{(0,1')\}; H:=\{(1,2),(1',2)\} $$