Question 1209282
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Here is one way to do a derivation using a conditional proof.
I'll use arrow symbols in place of horseshoe symbols.
<table border = "1" cellpadding = "5"><tr><td colspan=2>Number</td><td>Statement</td><td>Line(s) Used</td><td>Reason</td></tr><tr><td>1</td><td></td><td>~(~Z v H) --> ~T</td><td></td><td></td></tr><tr><td>2</td><td></td><td>(S --> Z) --> ~H</td><td></td><td></td></tr><tr><td>:.</td><td></td><td>Z --> ~T</td><td></td><td></td></tr><tr><td></td><td>3</td><td>Z</td><td></td><td>Assumption for Conditional Proof</td></tr><tr><td></td><td>4</td><td>(Z & ~H) --> ~T</td><td>1</td><td>De Morgan’s Law</td></tr><tr><td></td><td>5</td><td>Z --> (~H --> ~T)</td><td>4</td><td>Exportation</td></tr><tr><td></td><td>6</td><td>~H --> ~T</td><td>5,3</td><td>Modus Ponens</td></tr><tr><td></td><td>7</td><td>(S --> Z) --> ~T</td><td>2,6</td><td>Hypothetical Syllogism</td></tr><tr><td></td><td>8</td><td>(~S v Z) --> ~T</td><td>7</td><td>Material Implication</td></tr><tr><td></td><td>9</td><td>Z v ~S</td><td>3</td><td>Addition</td></tr><tr><td></td><td>10</td><td>~S v Z</td><td>9</td><td>Commutation</td></tr><tr><td></td><td>11</td><td>~T</td><td>8,10</td><td>Modus Ponens</td></tr><tr><td>12</td><td></td><td>Z --> ~T</td><td>3 - 11</td><td>Conditional Proof</td></tr></table>
I started with assuming Z is the case (line 3). Then I used the <a href="https://www.algebra.com/algebra/homework/Conjunction/logic-rules-of-inference-and-replacement.lesson">logic rules of inference and replacement</a> to arrive at ~T (line 11)


The assumption Z leading to ~T then allows us to prove Z --> ~T is a valid conclusion. 


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Here's a way to do it using a direct proof.
<table border = "1" cellpadding = "5"><tr><td>Number</td><td>Statement</td><td>Line(s) Used</td><td>Reason</td></tr><tr><td>1</td><td>~(~Z v H) --> ~T</td><td></td><td></td></tr><tr><td>2</td><td>(S --> Z) --> ~H</td><td></td><td></td></tr><tr><td>:.</td><td>Z --> ~T</td><td></td><td></td></tr><tr><td>3</td><td>(~S v Z) --> ~H</td><td>2</td><td>Material Implication</td></tr><tr><td>4</td><td>~(~S v Z) v ~H</td><td>3</td><td>Material Implication</td></tr><tr><td>5</td><td>(S & ~Z) v ~H</td><td>4</td><td>De Morgan’s Law</td></tr><tr><td>6</td><td>(S v ~H) & (~Z v ~H)</td><td>5</td><td>Distribution</td></tr><tr><td>7</td><td>(~Z v ~H) & (S v ~H)</td><td>6</td><td>Commutation</td></tr><tr><td>8</td><td>~Z v ~H</td><td>7</td><td>Simplification</td></tr><tr><td>9</td><td>Z --> ~H</td><td>8</td><td>Material Implication</td></tr><tr><td>10</td><td>(Z & ~H) --> ~T</td><td>1</td><td>De Morgan’s Law</td></tr><tr><td>11</td><td>(~H & Z) --> ~T</td><td>10</td><td>Commutation</td></tr><tr><td>12</td><td>~H --> (Z --> ~T)</td><td>11</td><td>Exportation</td></tr><tr><td>13</td><td>Z --> (Z --> ~T)</td><td>9, 12</td><td>Hypothetical Syllogism</td></tr><tr><td>14</td><td>(Z & Z) --> ~T</td><td>13</td><td>Exportation</td></tr><tr><td>15</td><td>Z --> ~T</td><td>14</td><td>Tautology</td></tr></table>
There might be a much more efficient pathway, but I'm not able to think of it right now.


Caution: Tutor Edwin makes the mistake of using the Addition Rule on part of an expression rather than the entire thing. 
It is NOT valid to go from ~(T ⊃ ~Z) to ~(T ⊃ (~Z <font color=red>v H</font>))
Notice in this <a href="https://www.algebra.com/algebra/homework/Conjunction/logic-rules-of-inference-and-replacement.lesson">rule set</a> the "addition" property is in the "rules of inference" sub-block. The rules of inference must apply to the entire line. In contrast a rule of replacement can apply to a portion of a line. 
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