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Theorem djussxp2 33121
Description: Stronger version of djussxp 5825. (Contributed by Thierry Arnoux, 23-Jun-2024.)
Assertion
Ref Expression
djussxp2 𝑘𝐴 ({𝑘} × 𝐵) ⊆ (𝐴 × 𝑘𝐴 𝐵)
Distinct variable group:   𝐴,𝑘
Allowed substitution hint:   𝐵(𝑘)

Proof of Theorem djussxp2
StepHypRef Expression
1 nfcv 2922 . . . 4 𝑘𝐴
2 nfiu1 4986 . . . 4 𝑘 𝑘𝐴 𝐵
31, 2nfxp 5688 . . 3 𝑘(𝐴 × 𝑘𝐴 𝐵)
43iunssf 5001 . 2 ( 𝑘𝐴 ({𝑘} × 𝐵) ⊆ (𝐴 × 𝑘𝐴 𝐵) ↔ ∀𝑘𝐴 ({𝑘} × 𝐵) ⊆ (𝐴 × 𝑘𝐴 𝐵))
5 snssi 4746 . . 3 (𝑘𝐴 → {𝑘} ⊆ 𝐴)
6 ssiun2 5006 . . 3 (𝑘𝐴𝐵 𝑘𝐴 𝐵)
7 xpss12 5670 . . 3 (({𝑘} ⊆ 𝐴𝐵 𝑘𝐴 𝐵) → ({𝑘} × 𝐵) ⊆ (𝐴 × 𝑘𝐴 𝐵))
85, 6, 7syl2anc 596 . 2 (𝑘𝐴 → ({𝑘} × 𝐵) ⊆ (𝐴 × 𝑘𝐴 𝐵))
94, 8mprgbir 3083 1 𝑘𝐴 ({𝑘} × 𝐵) ⊆ (𝐴 × 𝑘𝐴 𝐵)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wcel 2145  wss 3899  {csn 4584   ciun 4951   × cxp 5653
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ral 3077  df-rex 3087  df-v 3452  df-ss 3916  df-sn 4585  df-iun 4953  df-opab 5168  df-xp 5661
This theorem is used by:  2ndresdju  33122  gsumpart  33503
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