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The algorithm is (men ~ patterns, women ~ terms):

    while ∃ free man m who still has a woman w to propose to {
       w = first woman on m’s list to whom m has not yet proposed
       if w is free
         (m, w) become engaged
       else some pair (m', w) already exists
         if w prefers m to m'
            m' becomes free
           (m, w) become engaged 
         else
           (m', w) remain engaged

This assumes that there are M patterns and N terms and that M = N. 
If M < N, there will not be a match unless there is a wildcard or a remainder (see their sections below) and we should short-circuit unless there is one.

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In such case, the algo above would return an assignment where at least one pattern-term pair is 'married' despite not matching each other. Any such solution means the list of patterns we're matching can't be matched to the list of terms.

We can detect such solutions early, at the end of each iteration of the main loop. At the point where the algo finishes the main loop iteration for a given man m, there's no chance m will get a better option later. In fact, it can only get worse b/c of the swap that happens in lines 6-8. This means, in our case, that once a pattern ends its loop and doesn't have a matching term assigned, we won't be able to match the patterns to the terms and can terminate.

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