Why do we need 'seq' or 'pseq' with 'par' in Haskell?

Question

I\'m trying to understand why we need all parts of the standard sample code:

a `par` b `pseq` a+b

Why won\'t the following be sufficient?

Answer 1

a `par` b `par` a+b 

will evaluate a and b in parallel and returns a+b, yes.

However, the pseq there ensures both a and b are evaluated before a+b is.

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