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Theorem xppss12 42247
Description: Proper subset theorem for Cartesian product. (Contributed by Steven Nguyen, 17-Jul-2022.)
Assertion
Ref Expression
xppss12 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊊ (𝐵 × 𝐷))

Proof of Theorem xppss12
StepHypRef Expression
1 pssss 4108 . . 3 (𝐴𝐵𝐴𝐵)
2 pssss 4108 . . 3 (𝐶𝐷𝐶𝐷)
3 xpss12 5704 . . 3 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷))
41, 2, 3syl2an 596 . 2 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷))
5 simpl 482 . . . . 5 ((𝐴𝐵𝐶𝐷) → 𝐴𝐵)
6 pssne 4109 . . . . . 6 (𝐴𝐵𝐴𝐵)
76necomd 2994 . . . . 5 (𝐴𝐵𝐵𝐴)
8 neneq 2944 . . . . . 6 (𝐵𝐴 → ¬ 𝐵 = 𝐴)
98intnanrd 489 . . . . 5 (𝐵𝐴 → ¬ (𝐵 = 𝐴𝐷 = 𝐶))
105, 7, 93syl 18 . . . 4 ((𝐴𝐵𝐶𝐷) → ¬ (𝐵 = 𝐴𝐷 = 𝐶))
11 pssn0 42245 . . . . 5 (𝐴𝐵𝐵 ≠ ∅)
12 pssn0 42245 . . . . 5 (𝐶𝐷𝐷 ≠ ∅)
13 xp11 6197 . . . . 5 ((𝐵 ≠ ∅ ∧ 𝐷 ≠ ∅) → ((𝐵 × 𝐷) = (𝐴 × 𝐶) ↔ (𝐵 = 𝐴𝐷 = 𝐶)))
1411, 12, 13syl2an 596 . . . 4 ((𝐴𝐵𝐶𝐷) → ((𝐵 × 𝐷) = (𝐴 × 𝐶) ↔ (𝐵 = 𝐴𝐷 = 𝐶)))
1510, 14mtbird 325 . . 3 ((𝐴𝐵𝐶𝐷) → ¬ (𝐵 × 𝐷) = (𝐴 × 𝐶))
16 neqne 2946 . . . 4 (¬ (𝐵 × 𝐷) = (𝐴 × 𝐶) → (𝐵 × 𝐷) ≠ (𝐴 × 𝐶))
1716necomd 2994 . . 3 (¬ (𝐵 × 𝐷) = (𝐴 × 𝐶) → (𝐴 × 𝐶) ≠ (𝐵 × 𝐷))
1815, 17syl 17 . 2 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ≠ (𝐵 × 𝐷))
19 df-pss 3983 . 2 ((𝐴 × 𝐶) ⊊ (𝐵 × 𝐷) ↔ ((𝐴 × 𝐶) ⊆ (𝐵 × 𝐷) ∧ (𝐴 × 𝐶) ≠ (𝐵 × 𝐷)))
204, 18, 19sylanbrc 583 1 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊊ (𝐵 × 𝐷))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 206  wa 395   = wceq 1537  wne 2938  wss 3963  wpss 3964  c0 4339   × cxp 5687
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-11 2155  ax-12 2175  ax-ext 2706  ax-sep 5302  ax-nul 5312  ax-pr 5438
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ne 2939  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-ss 3980  df-pss 3983  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-br 5149  df-opab 5211  df-xp 5695  df-rel 5696  df-cnv 5697  df-dm 5699  df-rn 5700
This theorem is referenced by: (None)
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