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Theorem xppss12 42241
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 4046 . . 3 (𝐴𝐵𝐴𝐵)
2 pssss 4046 . . 3 (𝐶𝐷𝐶𝐷)
3 xpss12 5629 . . 3 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷))
41, 2, 3syl2an 596 . 2 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷))
5 simpl 482 . . . . 5 ((𝐴𝐵𝐶𝐷) → 𝐴𝐵)
6 pssne 4047 . . . . . 6 (𝐴𝐵𝐴𝐵)
76necomd 2981 . . . . 5 (𝐴𝐵𝐵𝐴)
8 neneq 2932 . . . . . 6 (𝐵𝐴 → ¬ 𝐵 = 𝐴)
98intnanrd 489 . . . . 5 (𝐵𝐴 → ¬ (𝐵 = 𝐴𝐷 = 𝐶))
105, 7, 93syl 18 . . . 4 ((𝐴𝐵𝐶𝐷) → ¬ (𝐵 = 𝐴𝐷 = 𝐶))
11 pssn0 42239 . . . . 5 (𝐴𝐵𝐵 ≠ ∅)
12 pssn0 42239 . . . . 5 (𝐶𝐷𝐷 ≠ ∅)
13 xp11 6119 . . . . 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 2934 . . . 4 (¬ (𝐵 × 𝐷) = (𝐴 × 𝐶) → (𝐵 × 𝐷) ≠ (𝐴 × 𝐶))
1716necomd 2981 . . 3 (¬ (𝐵 × 𝐷) = (𝐴 × 𝐶) → (𝐴 × 𝐶) ≠ (𝐵 × 𝐷))
1815, 17syl 17 . 2 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ≠ (𝐵 × 𝐷))
19 df-pss 3920 . 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 1541  wne 2926  wss 3900  wpss 3901  c0 4281   × cxp 5612
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2112  ax-9 2120  ax-11 2159  ax-12 2179  ax-ext 2702  ax-sep 5232  ax-nul 5242  ax-pr 5368
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2722  df-clel 2804  df-ne 2927  df-ral 3046  df-rex 3055  df-rab 3394  df-v 3436  df-dif 3903  df-un 3905  df-ss 3917  df-pss 3920  df-nul 4282  df-if 4474  df-sn 4575  df-pr 4577  df-op 4581  df-br 5090  df-opab 5152  df-xp 5620  df-rel 5621  df-cnv 5622  df-dm 5624  df-rn 5625
This theorem is referenced by: (None)
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