[+] A2 Finish latex
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@@ -111,7 +111,7 @@ $\forall s \in sc.sections,~ ...$
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% TODO: fill in the predicate definition for two meeting times conflicting
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\begin{align*}
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MeetingTimesConflict(m_1, m_2) : m_1[0] == m_2[0] \land m_1[2] > m_2[1] \land m_2[2] > m_1[1]
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MeetingTimesConflict(m_1, m_2) : m_1[0] == m_2[0] \land m_1[2] > m_2[1] \land m_2[2] > m_1[1] \\
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\qquad \text{where $m_1, m_2 \in M$}
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\end{align*}
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@@ -122,8 +122,8 @@ MeetingTimesConflict(m_1, m_2) : m_1[0] == m_2[0] \land m_1[2] > m_2[1] \land m_
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% Use the MeetingTimesConflict predicate in your response.
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\begin{align*}
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SectionsConflict(s_1, s_2) : ('Y' \in \{s_1[1], s_2[1]\} \lor s_1[1] = s_2[1]) \Rightarrow \\
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\exists m_1 \in s_1[2], \text{ s.t. } \exists m_2 \in s_2[2], \text { s.t. } MeetingTimesConflict(m_1, m_2)
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SectionsConflict(s_1, s_2) : (s_1[1]=\text{'Y'} \lor s_2[1]=\text{'Y'} \lor s_1[1] = s_2[1]) \land \\
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\exists m_1 \in s_1[2], \exists m_2 \in s_2[2], \text { s.t. } MeetingTimesConflict(m_1, m_2) \\
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\qquad \text{where $s_1, s_2 \in S$}
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\end{align*}
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