Theorem trreleq 39864

Description: Equality theorem for the transitive relation predicate. (Contributed by Peter Mazsa, 15-Apr-2019.) (Revised by Peter Mazsa, 23-Sep-2021.)

Assertion

Ref	Expression
trreleq	⊢ (𝑅 = 𝑆 → ( TrRel 𝑅 ↔ TrRel 𝑆))

Proof of Theorem trreleq

Step	Hyp	Ref	Expression
1		coideq 39446	. . . 4 ⊢ (𝑅 = 𝑆 → (𝑅 ∘ 𝑅) = (𝑆 ∘ 𝑆))
2		id 23	. . . 4 ⊢ (𝑅 = 𝑆 → 𝑅 = 𝑆)
3	1, 2	sseq12d 3946	. . 3 ⊢ (𝑅 = 𝑆 → ((𝑅 ∘ 𝑅) ⊆ 𝑅 ↔ (𝑆 ∘ 𝑆) ⊆ 𝑆))
4		releq 5761	. . 3 ⊢ (𝑅 = 𝑆 → (Rel 𝑅 ↔ Rel 𝑆))
5	3, 4	anbi12d 644	. 2 ⊢ (𝑅 = 𝑆 → (((𝑅 ∘ 𝑅) ⊆ 𝑅 ∧ Rel 𝑅) ↔ ((𝑆 ∘ 𝑆) ⊆ 𝑆 ∧ Rel 𝑆)))
6		dftrrel2 39859	. 2 ⊢ ( TrRel 𝑅 ↔ ((𝑅 ∘ 𝑅) ⊆ 𝑅 ∧ Rel 𝑅))
7		dftrrel2 39859	. 2 ⊢ ( TrRel 𝑆 ↔ ((𝑆 ∘ 𝑆) ⊆ 𝑆 ∧ Rel 𝑆))
8	5, 6, 7	3bitr4g 317	1 ⊢ (𝑅 = 𝑆 → ( TrRel 𝑅 ↔ TrRel 𝑆))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ⊆ wss 3900   ∘ ccom 5662  Rel wrel 5663   TrRel wtrrel 39396

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5255  ax-pr 5398

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-br 5109  df-opab 5174  df-xp 5664  df-rel 5665  df-cnv 5666  df-co 5667  df-dm 5668  df-rn 5669  df-res 5670  df-trrel 39856

This theorem is used by:  eqvreleq  39884