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Theorem relssres 4696
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 2613 . . . . . . . . 9 𝑥 ∈ V
3 vex 2613 . . . . . . . . 9 𝑦 ∈ V
42, 3opeldm 4586 . . . . . . . 8 (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥 ∈ dom 𝐴)
5 ssel 3002 . . . . . . . 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 4665 . . . . . 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 4478 . . 3 ((Rel 𝐴 ∧ dom 𝐴𝐵) → 𝐴 ⊆ (𝐴𝐵))
12 resss 4683 . . 3 (𝐴𝐵) ⊆ 𝐴
1311, 12jctil 305 . 2 ((Rel 𝐴 ∧ dom 𝐴𝐵) → ((𝐴𝐵) ⊆ 𝐴𝐴 ⊆ (𝐴𝐵)))
14 eqss 3023 . 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 1285  wcel 1434  wss 2982  cop 3419  dom cdm 4391  cres 4393  Rel wrel 4396
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 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-14 1446  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065  ax-sep 3916  ax-pow 3968  ax-pr 3992
This theorem depends on definitions:  df-bi 115  df-3an 922  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-ral 2358  df-rex 2359  df-v 2612  df-un 2986  df-in 2988  df-ss 2995  df-pw 3402  df-sn 3422  df-pr 3423  df-op 3425  df-br 3806  df-opab 3860  df-xp 4397  df-rel 4398  df-dm 4401  df-res 4403
This theorem is referenced by:  resdm  4697  resid  4712  fnresdm  5059  f1ompt  5373
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