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Theorem relpeq1 45389
Description: Equality theorem for relation-preserving functions. (Contributed by Eric Schmidt, 11-Oct-2025.)
Assertion
Ref Expression
relpeq1 (𝐻 = 𝐺 → (𝐻 RelPres 𝑅, 𝑆(𝐴, 𝐵) ↔ 𝐺 RelPres 𝑅, 𝑆(𝐴, 𝐵)))

Proof of Theorem relpeq1
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 feq1 6640 . . 3 (𝐻 = 𝐺 → (𝐻:𝐴𝐵𝐺:𝐴𝐵))
2 fveq1 6833 . . . . . 6 (𝐻 = 𝐺 → (𝐻𝑥) = (𝐺𝑥))
3 fveq1 6833 . . . . . 6 (𝐻 = 𝐺 → (𝐻𝑦) = (𝐺𝑦))
42, 3breq12d 5099 . . . . 5 (𝐻 = 𝐺 → ((𝐻𝑥)𝑆(𝐻𝑦) ↔ (𝐺𝑥)𝑆(𝐺𝑦)))
54imbi2d 340 . . . 4 (𝐻 = 𝐺 → ((𝑥𝑅𝑦 → (𝐻𝑥)𝑆(𝐻𝑦)) ↔ (𝑥𝑅𝑦 → (𝐺𝑥)𝑆(𝐺𝑦))))
652ralbidv 3202 . . 3 (𝐻 = 𝐺 → (∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦 → (𝐻𝑥)𝑆(𝐻𝑦)) ↔ ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦 → (𝐺𝑥)𝑆(𝐺𝑦))))
71, 6anbi12d 633 . 2 (𝐻 = 𝐺 → ((𝐻:𝐴𝐵 ∧ ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦 → (𝐻𝑥)𝑆(𝐻𝑦))) ↔ (𝐺:𝐴𝐵 ∧ ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦 → (𝐺𝑥)𝑆(𝐺𝑦)))))
8 df-relp 45388 . 2 (𝐻 RelPres 𝑅, 𝑆(𝐴, 𝐵) ↔ (𝐻:𝐴𝐵 ∧ ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦 → (𝐻𝑥)𝑆(𝐻𝑦))))
9 df-relp 45388 . 2 (𝐺 RelPres 𝑅, 𝑆(𝐴, 𝐵) ↔ (𝐺:𝐴𝐵 ∧ ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦 → (𝐺𝑥)𝑆(𝐺𝑦))))
107, 8, 93bitr4g 314 1 (𝐻 = 𝐺 → (𝐻 RelPres 𝑅, 𝑆(𝐴, 𝐵) ↔ 𝐺 RelPres 𝑅, 𝑆(𝐴, 𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1542  wral 3052   class class class wbr 5086  wf 6488  cfv 6492   RelPres wrelp 45387
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rab 3391  df-v 3432  df-dif 3893  df-un 3895  df-ss 3907  df-nul 4275  df-if 4468  df-sn 4569  df-pr 4571  df-op 4575  df-uni 4852  df-br 5087  df-opab 5149  df-rel 5631  df-cnv 5632  df-co 5633  df-dm 5634  df-rn 5635  df-iota 6448  df-fun 6494  df-fn 6495  df-f 6496  df-fv 6500  df-relp 45388
This theorem is referenced by: (None)
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