Theorem relcnvtrg 6268

Description: Subclass law for converse of a composition. A particular case is ((𝑅 ∘ 𝑅) ⊆ 𝑅 ↔ (^◡𝑅 ∘ ^◡𝑅) ⊆ ^◡𝑅), which says that a relation is transitive iff its converse is transitive. (Contributed by FL, 19-Sep-2011.) Generalize to a statement about three classes rather than one. (Revised by Peter Mazsa, 17-Oct-2023.) Remove antecedent. (Revised by Eric Schmidt, 8-Aug-2026.)

Assertion

Ref	Expression
relcnvtrg	⊢ ((𝑅 ∘ 𝑆) ⊆ 𝑇 ↔ (◡𝑆 ∘ ◡𝑅) ⊆ ◡𝑇)

Proof of Theorem relcnvtrg

Step	Hyp	Ref	Expression
1		cnvco 5866	. . 3 ⊢ ◡(𝑅 ∘ 𝑆) = (◡𝑆 ∘ ◡𝑅)
2		cnvss 5849	. . 3 ⊢ ((𝑅 ∘ 𝑆) ⊆ 𝑇 → ◡(𝑅 ∘ 𝑆) ⊆ ◡𝑇)
3	1, 2	eqsstrrid 3947	. 2 ⊢ ((𝑅 ∘ 𝑆) ⊆ 𝑇 → (◡𝑆 ∘ ◡𝑅) ⊆ ◡𝑇)
4		cnvco 5866	. . . . 5 ⊢ ◡(◡𝑆 ∘ ◡𝑅) = (◡◡𝑅 ∘ ◡◡𝑆)
5		cocnvcnv1 6259	. . . . 5 ⊢ (◡◡𝑅 ∘ ◡◡𝑆) = (𝑅 ∘ ◡◡𝑆)
6		cocnvcnv2 6260	. . . . 5 ⊢ (𝑅 ∘ ◡◡𝑆) = (𝑅 ∘ 𝑆)
7	4, 5, 6	3eqtri 2787	. . . 4 ⊢ ◡(◡𝑆 ∘ ◡𝑅) = (𝑅 ∘ 𝑆)
8		cnvss 5849	. . . 4 ⊢ ((◡𝑆 ∘ ◡𝑅) ⊆ ◡𝑇 → ◡(◡𝑆 ∘ ◡𝑅) ⊆ ◡◡𝑇)
9	7, 8	eqsstrrid 3947	. . 3 ⊢ ((◡𝑆 ∘ ◡𝑅) ⊆ ◡𝑇 → (𝑅 ∘ 𝑆) ⊆ ◡◡𝑇)
10		cnvcnvss 6185	. . 3 ⊢ ◡◡𝑇 ⊆ 𝑇
11	9, 10	sstrdi 3920	. 2 ⊢ ((◡𝑆 ∘ ◡𝑅) ⊆ ◡𝑇 → (𝑅 ∘ 𝑆) ⊆ 𝑇)
12	3, 11	impbii 212	1 ⊢ ((𝑅 ∘ 𝑆) ⊆ 𝑇 ↔ (◡𝑆 ∘ ◡𝑅) ⊆ ◡𝑇)

This proof depends on syntax axioms:   ↔ wb 209   ⊆ wss 3895  ^◡ccnv 5649   ∘ ccom 5654

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248  ax-pr 5390

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-br 5103  df-opab 5167  df-xp 5656  df-rel 5657  df-cnv 5658  df-co 5659  df-dm 5660  df-rn 5661  df-res 5662

This theorem is used by: (None)