Considérons V Z = {z j , z k }, d'après le Théorème 5.4.3, V Z est une solution pour leprobì eme de RPRM. Dans ce cas, la suppression de z j ne change pas la solubilité duprobì eme de RPRM parce que {V Z \z j } = {z k } est une solution ,
nous avons I * = {x 2 } et la perturbation d affecte seulement x 2 ? I * . D'après la Définition 5.4.2, nous trouvons D d = ?. Sans aucune mesure, on a J * = {x 1 } donc I * ? J * = X alors leprobì eme de RPRM est soluble sans aucune mesure, En effet ,
nous avons I * = {x 1 ,
x 8 }, alors leprobì eme RPRM est soluble d'après le Théorème 5.2.5. Nous avons F I * = {x 2 ,
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