Simple Proof Question on Fundamentals (if x implies y, and y implies z, how does x imply z?) So as the title says, the question I am attempting to wrap by head around is "x implies y, y implies z, then x implies z". It seemed almost like a joke, I thought the answer was right in the question.
Assume $x \Rightarrow y$
Assume $y \Rightarrow z$
Then $x \Rightarrow y$
Then $y \Rightarrow z$
Therefore, $x \Rightarrow z$
So it seems real simple, right? I showed my friend, but he says there is more to it. Can someone help me figure out what I am missing? Because on initial glance, the question seems too easy to be true...
 A: To prove $x \implies z$, it suffices to show that if $x$ is true, then $z$ must be true.
If $x$ is true, from $x \implies y$, we know that $y$ is true.
Now, from $y \implies z$ and $y$ is true, we conclude that $z$ is true.
Hence $x \implies z$
A: The statement (x implies y) is defined to be the statement ((not x) or y). Intuitively, this makes sense: since x implies y, it should not happen that both x and (not y) hold at the same time.
Now we prove: if x implies y and y implies z, then x implies z.
If (not x) holds, then we are done, since then ((not x) or z) must hold. 
If (not x) does not hold, then y must hold, since x implies y. Hence, (not y) does not hold. Hence z holds, by the definition of (y implies z). So we are done in this case also.
Since there are no more possible cases, we are done.
If this argument does not convince you, then you could try to draw the truth table for the possible values of x,y,z.
A: Here are three ways to see that if $X\to Y$ and $Y\to Z$ then $X\to Z$.


*

*Consider how this might look in a diagram. If $X$ is a sufficient condition for something else, say, $Y$, then whenever we have $X$ we have $Y$. That could be represented as one circle containing everything in $X$ being fully contained in a circle containing everything in $Y$.  Here is how those relationships between $X$, $Y$ and $Z$ might be drawn:





*A natural deduction proof may help clarify the inference rules needed to go from the premises, $X\to Y$ and $Y\to Z$, to the goal, $X\to Z$. Here is a Fitch-style natural deduction proof using only two inference rules, implication elimination ($\to$E) and implication introduction ($\to$I):





*If one conjoins the two premises into one proposition, $(X\to Y)\land(Y\to Z)$, and then writes that as the antecedent of an implication leading to the consequent, $X\to Z$, one can show that the resulting proposition is a tautology by using a truth table. Note that the column in red has only the value "T" standing for true. For all valuations of $X$, $Y$ and $Z$ this top level connection between premises and conclusion is true.




Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
A: Yes, just a bit more is needed.


*

*Assume $x\to y$ as a premise.

*Assume $y\to z$ as a premise.

*Assume $x$.

*Derive $y$ from 1 and 3 via modus ponens.

*Derive $z$ from 2 and 4 via modus ponens.

*Deduce $x\to z$ from subproof 3-5.
$\blacksquare~x\to z$ is derivable from $x\to y, y\to z$ 
A: Not according to this (Question 16), it shows that if we assume that $X\rightarrow Y$ and $Y\rightarrow Z$, and if X is false then no conclusion can be drawn, I would expect both $Y$ and $Z$ to be false as well.
