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In order to study the congruence relation $\\eta^{\\ast}$ on finite semigroups, we define a $\\textbf{CS}$-diagonal finite regular Rees matrix semigroup. We prove that, if $S$ is a $\\textbf{CS}$-diagonal finite regular Rees matrix semigroup then $S/\\eta^{\\ast}$ is inverse. Also, if $S$ is a completely regular finite semigroup, then $S/\\eta^{\\ast}$ is a Clifford semigroup.\n  We"},"verification_status":{"content_addressed":true,"pith_receipt":true,"author_attested":false,"weak_author_claims":0,"strong_author_claims":0,"externally_anchored":false,"storage_verified":false,"citation_signatures":0,"replication_records":0,"graph_snapshot":true,"references_resolved":false,"formal_links_present":false},"canonical_record":{"source":{"id":"1410.2194","kind":"arxiv","version":2},"metadata":{"license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","primary_cat":"math.GR","submitted_at":"2014-10-08T17:33:28Z","cross_cats_sorted":[],"title_canon_sha256":"bff89da6947c74d4c7d24f3e4cfc466cc52c7d07795b72a02c135657350bf02d","abstract_canon_sha256":"f44e31a2640af37a4453f2fd30b35aa4482548556ae33af17835e549ac5bafa6"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-18T01:11:33.738918Z","signature_b64":"DNrAlsGo0IMKNvvR/1ZW9W1mpEhR01lSJ1qsBeTqs1xYTdpfkxAfRRBq5KV7zf445s69d6qnycd7a6zA3KC1Ag==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"39ceaa10a7226be3b4f074ee4c14d99f7af8a7ebf57716617517be5a1c0c108b","last_reissued_at":"2026-05-18T01:11:33.738559Z","signature_status":"signed_v1","first_computed_at":"2026-05-18T01:11:33.738559Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"The congruence $\\eta^{\\ast}$ on semigroups","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.GR","authors_text":"M.H. 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