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Theorem relssres 4675
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 106 . . . 4 ((Rel 𝐴 ∧ dom 𝐴𝐵) → Rel 𝐴)
2 vex 2577 . . . . . . . . 9 𝑥 ∈ V
3 vex 2577 . . . . . . . . 9 𝑦 ∈ V
42, 3opeldm 4565 . . . . . . . 8 (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥 ∈ dom 𝐴)
5 ssel 2966 . . . . . . . 8 (dom 𝐴𝐵 → (𝑥 ∈ dom 𝐴𝑥𝐵))
64, 5syl5 32 . . . . . . 7 (dom 𝐴𝐵 → (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥𝐵))
76ancld 312 . . . . . 6 (dom 𝐴𝐵 → (⟨𝑥, 𝑦⟩ ∈ 𝐴 → (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥𝐵)))
83opelres 4644 . . . . . 6 (⟨𝑥, 𝑦⟩ ∈ (𝐴𝐵) ↔ (⟨𝑥, 𝑦⟩ ∈ 𝐴𝑥𝐵))
97, 8syl6ibr 155 . . . . 5 (dom 𝐴𝐵 → (⟨𝑥, 𝑦⟩ ∈ 𝐴 → ⟨𝑥, 𝑦⟩ ∈ (𝐴𝐵)))
109adantl 266 . . . 4 ((Rel 𝐴 ∧ dom 𝐴𝐵) → (⟨𝑥, 𝑦⟩ ∈ 𝐴 → ⟨𝑥, 𝑦⟩ ∈ (𝐴𝐵)))
111, 10relssdv 4459 . . 3 ((Rel 𝐴 ∧ dom 𝐴𝐵) → 𝐴 ⊆ (𝐴𝐵))
12 resss 4662 . . 3 (𝐴𝐵) ⊆ 𝐴
1311, 12jctil 299 . 2 ((Rel 𝐴 ∧ dom 𝐴𝐵) → ((𝐴𝐵) ⊆ 𝐴𝐴 ⊆ (𝐴𝐵)))
14 eqss 2987 . 2 ((𝐴𝐵) = 𝐴 ↔ ((𝐴𝐵) ⊆ 𝐴𝐴 ⊆ (𝐴𝐵)))
1513, 14sylibr 141 1 ((Rel 𝐴 ∧ dom 𝐴𝐵) → (𝐴𝐵) = 𝐴)
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101   = wceq 1259  wcel 1409  wss 2944  cop 3405  dom cdm 4372  cres 4374  Rel wrel 4377
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-14 1421  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038  ax-sep 3902  ax-pow 3954  ax-pr 3971
This theorem depends on definitions:  df-bi 114  df-3an 898  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-ral 2328  df-rex 2329  df-v 2576  df-un 2949  df-in 2951  df-ss 2958  df-pw 3388  df-sn 3408  df-pr 3409  df-op 3411  df-br 3792  df-opab 3846  df-xp 4378  df-rel 4379  df-dm 4382  df-res 4384
This theorem is referenced by:  resdm  4676  resid  4689  fnresdm  5035  f1ompt  5347
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