What is the difference between strong normalization and weak normalization in the context of rewrite systems?

In the context of rewriting systems, how does strong normalization differ from weak normalization?

• The article you linked literally answers your question. Dec 30 '16 at 23:10
• I know it does, but I was having difficulty realising how the two kinds of normalization play together, which @chi kindly solved below. It is sometimes good practice (and a token of empathic thinking) to answer to the need behind the question than only take the question in its literal form. Dec 30 '16 at 23:21
• It would likewise be a kind gesture of the asker to explain in sufficient detail what would be expected from an answer, rather than expecting others to read his mind. Dec 30 '16 at 23:27
• @quicksort Some people are worse than others in reading the crux of a simple question. Downvote and move on please. Dec 30 '16 at 23:32
• So do not make it harder for them. quicksort is right, it would be nice to reveal the expectations. Just because the mind reading succeded this time it will probably not be the case the most of the time. No downvote, sorry.
– Evil
Dec 30 '16 at 23:46

$$S \rightarrow (SS) \qquad S \rightarrow A$$
is weakly normalizing (given any term, we can rewrite every $S$ into an $A$, after which normal form is reached) but not strongly normalizing ($S \rightarrow (SS) \rightarrow ((S S) S) \rightarrow \cdots$).