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Drag 8 statements from the left to the right side below, and arrange them them in the correct order to produce a proof of the following proposition.

Proposition. Let $\mathbb{Z}-\lbrace 0 \rbrace$ denote the set of nonzero integers. Define a relation $R$ on $A = \mathbb{Z} \times (\mathbb{Z}-\lbrace 0 \rbrace)$ by:
$((a,b),(c,d)) \in R \iff ad=bc$ Then $R$ is an equivalence relation on the set $A$ .

Choose from these statements

Therefore $R$ is symmetric.
Next suppose $(a,b) R (c,d)$ and $(c,d) R (e,f)$ . Then $ad = bc$ and $cf=de$ and
Define $R$ on $\mathbb{Z} \times \mathbb{Z}$ such that $((a,b),(c,d)) \in R \iff ab=cd$ .
$b,d,f$ are nonzero, so
$adf=bcf=bde$ .
Cancelling $d$ we obtain
Hence, $R$ is symmetric since $((a,b),(a,b)) \in R$ .
Hence $R$ is a linear ordering because $a>b \ \& \ b>c \implies a>c$ .
and so $(a,b) R (a,b)$ , so $R$ is reflexive.
Then $ad=bc$ , so $bc =ad$ , hence $(c,d) R (a,b)$ .
Nathan is a goob.
Therefore, $R$ is transitive.
Obviously $ab = ba$ whenever $(a,b)\in R$ ,
Therefore $R$ is an equivalence relation.
$af=be$ .
Thus $(a,b) R (e,f)$ because $b,f \neq 0$ .
$(a,b) R (c,d) \implies ab = cd$ .
Hence, $R$ is reflexive and $(a,b) R (c,d)$ means $R$ is symmetric and transitive.
Suppose $((a,b),(c,d)) \in R$ .

Proof: (A logical proof of the given proposition.)

Choose from these statements

Therefore $R$ is symmetric.
Next suppose $(a,b) R (c,d)$ and $(c,d) R (e,f)$ . Then $ad = bc$ and $cf=de$ and
Define $R$ on $\mathbb{Z} \times \mathbb{Z}$ such that $((a,b),(c,d)) \in R \iff ab=cd$ .
$b,d,f$ are nonzero, so
$adf=bcf=bde$ .
Cancelling $d$ we obtain
Hence, $R$ is symmetric since $((a,b),(a,b)) \in R$ .
Hence $R$ is a linear ordering because $a>b \ \& \ b>c \implies a>c$ .
and so $(a,b) R (a,b)$ , so $R$ is reflexive.
Then $ad=bc$ , so $bc =ad$ , hence $(c,d) R (a,b)$ .
Nathan is a goob.
Therefore, $R$ is transitive.
Obviously $ab = ba$ whenever $(a,b)\in R$ ,
Therefore $R$ is an equivalence relation.
$af=be$ .
Thus $(a,b) R (e,f)$ because $b,f \neq 0$ .
$(a,b) R (c,d) \implies ab = cd$ .
Hence, $R$ is reflexive and $(a,b) R (c,d)$ means $R$ is symmetric and transitive.
Suppose $((a,b),(c,d)) \in R$ .

Proof: (A logical proof of the given proposition.)