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Theorem xppss12 42224
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 4064 . . 3 (𝐴𝐵𝐴𝐵)
2 pssss 4064 . . 3 (𝐶𝐷𝐶𝐷)
3 xpss12 5656 . . 3 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷))
41, 2, 3syl2an 596 . 2 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷))
5 simpl 482 . . . . 5 ((𝐴𝐵𝐶𝐷) → 𝐴𝐵)
6 pssne 4065 . . . . . 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 42222 . . . . 5 (𝐴𝐵𝐵 ≠ ∅)
12 pssn0 42222 . . . . 5 (𝐶𝐷𝐷 ≠ ∅)
13 xp11 6151 . . . . 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 3937 . 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 1540  wne 2926  wss 3917  wpss 3918  c0 4299   × cxp 5639
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-11 2158  ax-12 2178  ax-ext 2702  ax-sep 5254  ax-nul 5264  ax-pr 5390
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-ne 2927  df-ral 3046  df-rex 3055  df-rab 3409  df-v 3452  df-dif 3920  df-un 3922  df-ss 3934  df-pss 3937  df-nul 4300  df-if 4492  df-sn 4593  df-pr 4595  df-op 4599  df-br 5111  df-opab 5173  df-xp 5647  df-rel 5648  df-cnv 5649  df-dm 5651  df-rn 5652
This theorem is referenced by: (None)
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