Reduction from Clique to IS degree at most 4 This is from the Algorithms textbook by Dasgupta,C. H.Papadimitriou,andU. V. Vazirani question 8.6 (b) that asks:

Edit:

And missing from the pic as @Nathaniel points out in his answer below:
&quot;3SAT remains NP-complete even when restricted to formulas in which
each litteral appears at most twice&quot;.

My reduction from Clique:
Definition of clique: fully connected graph/subgraph
So given G=(V,E) and a goal g it returns a clique of size g fully connected.
The complement graph G’ would contain an IS of size g of the same vertices.
So given a graph G, with degree at most 4, as input to Clique, the largest possible clique would be of size 5 vertices per the  definition.  Let's assume g=5 as input to the Clique problem.
Make the complement graph G’ =(V, E’) where uv in E’ not in E and call ISdeg4 on G’ with g=5.
If there is an IS of size g=5 in G’, then those 5 vertices make up a clique in G so return that to Clique.
Else returns no which I can return to Clique.
ISdeg4 is in NP:
Returns a set of vertices which are an IS.
In O(n+m) I can check that none of the v's in the IS have an edge to another v in the IS.
So ISdeg4 can be used to solve Clique problems for G=(V,E) with degree at most 4.
Edit:
The above is not a solution since the reduction from Clique to ISDeg4 only solves a subset of Clique problems where an input G=(V,E) is at most degree 4.
Thus, I want to know if anyone has a reduction from a known NP-Complete graph problem rather than 3SAT-&gt; IsDeg4 which I was able to understand after seeing the solution given by my teacher.
As you stated, this is not a correct reduction, because you would need to solve Clique for any graph (or prove that Clique with graphs of degree at most 4 is $NP$-complete).
Actually, the question 8.6.b stands true if the graph have degree at most 3. I think the intention of the author for solving this exercice is to use the remark stated at the beginning of the exercice (which you cropped): &quot;$\textsf{3SAT}$ remains $\textsf{NP}$-complete even when restricted to formulas in which each litteral appears at most twice&quot;.
You should then make a reduction from $\textsf{3SAT-2LITmax}$ (a convenient name I came up for the problem I just described), using the same kind of reduction to prove $\textsf{NP}$-hardness of $\textsf{IS}$ from $\textsf{3SAT}$.