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Theorem relssres 4666
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 107 . . . 4 ((Rel 𝐴 ∧ dom 𝐴𝐵) → Rel 𝐴)
2 vex 2604 . . . . . . . . 9 𝑥 ∈ V
3 vex 2604 . . . . . . . . 9 𝑦 ∈ V
42, 3opeldm 4556 . . . . . . . 8 (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥 ∈ dom 𝐴)
5 ssel 2993 . . . . . . . 8 (dom 𝐴𝐵 → (𝑥 ∈ dom 𝐴𝑥𝐵))
64, 5syl5 32 . . . . . . 7 (dom 𝐴𝐵 → (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥𝐵))
76ancld 318 . . . . . 6 (dom 𝐴𝐵 → (⟨𝑥, 𝑦⟩ ∈ 𝐴 → (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥𝐵)))
83opelres 4635 . . . . . 6 (⟨𝑥, 𝑦⟩ ∈ (𝐴𝐵) ↔ (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥𝐵))
97, 8syl6ibr 160 . . . . 5 (dom 𝐴𝐵 → (⟨𝑥, 𝑦⟩ ∈ 𝐴 → ⟨𝑥, 𝑦⟩ ∈ (𝐴𝐵)))
109adantl 271 . . . 4 ((Rel 𝐴 ∧ dom 𝐴𝐵) → (⟨𝑥, 𝑦⟩ ∈ 𝐴 → ⟨𝑥, 𝑦⟩ ∈ (𝐴𝐵)))
111, 10relssdv 4450 . . 3 ((Rel 𝐴 ∧ dom 𝐴𝐵) → 𝐴 ⊆ (𝐴𝐵))
12 resss 4653 . . 3 (𝐴𝐵) ⊆ 𝐴
1311, 12jctil 305 . 2 ((Rel 𝐴 ∧ dom 𝐴𝐵) → ((𝐴𝐵) ⊆ 𝐴𝐴 ⊆ (𝐴𝐵)))
14 eqss 3014 . 2 ((𝐴𝐵) = 𝐴 ↔ ((𝐴𝐵) ⊆ 𝐴𝐴 ⊆ (𝐴𝐵)))
1513, 14sylibr 132 1 ((Rel 𝐴 ∧ dom 𝐴𝐵) → (𝐴𝐵) = 𝐴)
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102   = wceq 1284  wcel 1433  wss 2973  cop 3401  dom cdm 4363  cres 4365  Rel wrel 4368
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-14 1445  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063  ax-sep 3896  ax-pow 3948  ax-pr 3964
This theorem depends on definitions:  df-bi 115  df-3an 921  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-rex 2354  df-v 2603  df-un 2977  df-in 2979  df-ss 2986  df-pw 3384  df-sn 3404  df-pr 3405  df-op 3407  df-br 3786  df-opab 3840  df-xp 4369  df-rel 4370  df-dm 4373  df-res 4375
This theorem is referenced by:  resdm  4667  resid  4682  fnresdm  5028  f1ompt  5341
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