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Theorem sspred 6135
 Description: Another subset/predecessor class relationship. (Contributed by Scott Fenton, 6-Feb-2011.)
Assertion
Ref Expression
sspred ((𝐵𝐴 ∧ Pred(𝑅, 𝐴, 𝑋) ⊆ 𝐵) → Pred(𝑅, 𝐴, 𝑋) = Pred(𝑅, 𝐵, 𝑋))

Proof of Theorem sspred
StepHypRef Expression
1 sseqin2 4121 . 2 (𝐵𝐴 ↔ (𝐴𝐵) = 𝐵)
2 df-pred 6127 . . . 4 Pred(𝑅, 𝐴, 𝑋) = (𝐴 ∩ (𝑅 “ {𝑋}))
32sseq1i 3921 . . 3 (Pred(𝑅, 𝐴, 𝑋) ⊆ 𝐵 ↔ (𝐴 ∩ (𝑅 “ {𝑋})) ⊆ 𝐵)
4 df-ss 3876 . . 3 ((𝐴 ∩ (𝑅 “ {𝑋})) ⊆ 𝐵 ↔ ((𝐴 ∩ (𝑅 “ {𝑋})) ∩ 𝐵) = (𝐴 ∩ (𝑅 “ {𝑋})))
5 in32 4127 . . . 4 ((𝐴 ∩ (𝑅 “ {𝑋})) ∩ 𝐵) = ((𝐴𝐵) ∩ (𝑅 “ {𝑋}))
65eqeq1i 2764 . . 3 (((𝐴 ∩ (𝑅 “ {𝑋})) ∩ 𝐵) = (𝐴 ∩ (𝑅 “ {𝑋})) ↔ ((𝐴𝐵) ∩ (𝑅 “ {𝑋})) = (𝐴 ∩ (𝑅 “ {𝑋})))
73, 4, 63bitri 301 . 2 (Pred(𝑅, 𝐴, 𝑋) ⊆ 𝐵 ↔ ((𝐴𝐵) ∩ (𝑅 “ {𝑋})) = (𝐴 ∩ (𝑅 “ {𝑋})))
8 ineq1 4110 . . . . . 6 ((𝐴𝐵) = 𝐵 → ((𝐴𝐵) ∩ (𝑅 “ {𝑋})) = (𝐵 ∩ (𝑅 “ {𝑋})))
98eqeq1d 2761 . . . . 5 ((𝐴𝐵) = 𝐵 → (((𝐴𝐵) ∩ (𝑅 “ {𝑋})) = (𝐴 ∩ (𝑅 “ {𝑋})) ↔ (𝐵 ∩ (𝑅 “ {𝑋})) = (𝐴 ∩ (𝑅 “ {𝑋}))))
109biimpa 481 . . . 4 (((𝐴𝐵) = 𝐵 ∧ ((𝐴𝐵) ∩ (𝑅 “ {𝑋})) = (𝐴 ∩ (𝑅 “ {𝑋}))) → (𝐵 ∩ (𝑅 “ {𝑋})) = (𝐴 ∩ (𝑅 “ {𝑋})))
11 df-pred 6127 . . . 4 Pred(𝑅, 𝐵, 𝑋) = (𝐵 ∩ (𝑅 “ {𝑋}))
1210, 11, 23eqtr4g 2819 . . 3 (((𝐴𝐵) = 𝐵 ∧ ((𝐴𝐵) ∩ (𝑅 “ {𝑋})) = (𝐴 ∩ (𝑅 “ {𝑋}))) → Pred(𝑅, 𝐵, 𝑋) = Pred(𝑅, 𝐴, 𝑋))
1312eqcomd 2765 . 2 (((𝐴𝐵) = 𝐵 ∧ ((𝐴𝐵) ∩ (𝑅 “ {𝑋})) = (𝐴 ∩ (𝑅 “ {𝑋}))) → Pred(𝑅, 𝐴, 𝑋) = Pred(𝑅, 𝐵, 𝑋))
141, 7, 13syl2anb 601 1 ((𝐵𝐴 ∧ Pred(𝑅, 𝐴, 𝑋) ⊆ 𝐵) → Pred(𝑅, 𝐴, 𝑋) = Pred(𝑅, 𝐵, 𝑋))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ wa 400   = wceq 1539   ∩ cin 3858   ⊆ wss 3859  {csn 4523  ◡ccnv 5524   “ cima 5528  Predcpred 6126 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2114  ax-9 2122  ax-ext 2730 This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1542  df-ex 1783  df-sb 2071  df-clab 2737  df-cleq 2751  df-clel 2831  df-rab 3080  df-v 3412  df-in 3866  df-ss 3876  df-pred 6127 This theorem is referenced by:  frmin  33327
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