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Theorem relssres 4648
 Description: Simplification law for restriction. (Contributed by NM, 16-Aug-1994.)
Assertion
Ref Expression
relssres ((Rel 𝐴 ∧ dom 𝐴𝐵) → (𝐴𝐵) = 𝐴)

Proof of Theorem relssres
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 simpl 102 . . . 4 ((Rel 𝐴 ∧ dom 𝐴𝐵) → Rel 𝐴)
2 vex 2560 . . . . . . . . 9 𝑥 ∈ V
3 vex 2560 . . . . . . . . 9 𝑦 ∈ V
42, 3opeldm 4538 . . . . . . . 8 (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥 ∈ dom 𝐴)
5 ssel 2939 . . . . . . . 8 (dom 𝐴𝐵 → (𝑥 ∈ dom 𝐴𝑥𝐵))
64, 5syl5 28 . . . . . . 7 (dom 𝐴𝐵 → (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥𝐵))
76ancld 308 . . . . . 6 (dom 𝐴𝐵 → (⟨𝑥, 𝑦⟩ ∈ 𝐴 → (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥𝐵)))
83opelres 4617 . . . . . 6 (⟨𝑥, 𝑦⟩ ∈ (𝐴𝐵) ↔ (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥𝐵))
97, 8syl6ibr 151 . . . . 5 (dom 𝐴𝐵 → (⟨𝑥, 𝑦⟩ ∈ 𝐴 → ⟨𝑥, 𝑦⟩ ∈ (𝐴𝐵)))
109adantl 262 . . . 4 ((Rel 𝐴 ∧ dom 𝐴𝐵) → (⟨𝑥, 𝑦⟩ ∈ 𝐴 → ⟨𝑥, 𝑦⟩ ∈ (𝐴𝐵)))
111, 10relssdv 4432 . . 3 ((Rel 𝐴 ∧ dom 𝐴𝐵) → 𝐴 ⊆ (𝐴𝐵))
12 resss 4635 . . 3 (𝐴𝐵) ⊆ 𝐴
1311, 12jctil 295 . 2 ((Rel 𝐴 ∧ dom 𝐴𝐵) → ((𝐴𝐵) ⊆ 𝐴𝐴 ⊆ (𝐴𝐵)))
14 eqss 2960 . 2 ((𝐴𝐵) = 𝐴 ↔ ((𝐴𝐵) ⊆ 𝐴𝐴 ⊆ (𝐴𝐵)))
1513, 14sylibr 137 1 ((Rel 𝐴 ∧ dom 𝐴𝐵) → (𝐴𝐵) = 𝐴)
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 97   = wceq 1243   ∈ wcel 1393   ⊆ wss 2917  ⟨cop 3378  dom cdm 4345   ↾ cres 4347  Rel wrel 4350 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-14 1405  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022  ax-sep 3875  ax-pow 3927  ax-pr 3944 This theorem depends on definitions:  df-bi 110  df-3an 887  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311  df-rex 2312  df-v 2559  df-un 2922  df-in 2924  df-ss 2931  df-pw 3361  df-sn 3381  df-pr 3382  df-op 3384  df-br 3765  df-opab 3819  df-xp 4351  df-rel 4352  df-dm 4355  df-res 4357 This theorem is referenced by:  resdm  4649  resid  4662  fnresdm  5008  f1ompt  5320
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