Let witness that is measurable. Regularity is given, so it remains to prove the strong limit cardinal property. First, every has cardinality : if , thenby nonprincipality and -completeness, contradicting .
Suppose and . Choose an injection . For each , exactly one oflies in . Their chosen intersection lies in by -completeness. On that intersection every is the same subset of , contradicting injectivity because every member of has size . Thus , and is strongly inaccessible.
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