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Theorem xppss12 42860
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 4054 . . 3 (𝐴𝐵𝐴𝐵)
2 pssss 4054 . . 3 (𝐶𝐷𝐶𝐷)
3 xpss12 5667 . . 3 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷))
41, 2, 3syl2an 607 . 2 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷))
5 simpl 487 . . . . 5 ((𝐴𝐵𝐶𝐷) → 𝐴𝐵)
6 pssne 4055 . . . . . 6 (𝐴𝐵𝐴𝐵)
76necomd 3015 . . . . 5 (𝐴𝐵𝐵𝐴)
8 neneq 2966 . . . . . 6 (𝐵𝐴 → ¬ 𝐵 = 𝐴)
98intnanrd 494 . . . . 5 (𝐵𝐴 → ¬ (𝐵 = 𝐴𝐷 = 𝐶))
105, 7, 93syl 19 . . . 4 ((𝐴𝐵𝐶𝐷) → ¬ (𝐵 = 𝐴𝐷 = 𝐶))
11 pssn0 42858 . . . . 5 (𝐴𝐵𝐵 ≠ ∅)
12 pssn0 42858 . . . . 5 (𝐶𝐷𝐷 ≠ ∅)
13 xp11 6165 . . . . 5 ((𝐵 ≠ ∅ ∧ 𝐷 ≠ ∅) → ((𝐵 × 𝐷) = (𝐴 × 𝐶) ↔ (𝐵 = 𝐴𝐷 = 𝐶)))
1411, 12, 13syl2an 607 . . . 4 ((𝐴𝐵𝐶𝐷) → ((𝐵 × 𝐷) = (𝐴 × 𝐶) ↔ (𝐵 = 𝐴𝐷 = 𝐶)))
1510, 14mtbird 328 . . 3 ((𝐴𝐵𝐶𝐷) → ¬ (𝐵 × 𝐷) = (𝐴 × 𝐶))
16 neqne 2968 . . . 4 (¬ (𝐵 × 𝐷) = (𝐴 × 𝐶) → (𝐵 × 𝐷) ≠ (𝐴 × 𝐶))
1716necomd 3015 . . 3 (¬ (𝐵 × 𝐷) = (𝐴 × 𝐶) → (𝐴 × 𝐶) ≠ (𝐵 × 𝐷))
1815, 17syl 18 . 2 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ≠ (𝐵 × 𝐷))
19 df-pss 3927 . 2 ((𝐴 × 𝐶) ⊊ (𝐵 × 𝐷) ↔ ((𝐴 × 𝐶) ⊆ (𝐵 × 𝐷) ∧ (𝐴 × 𝐶) ≠ (𝐵 × 𝐷)))
204, 18, 19sylanbrc 594 1 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊊ (𝐵 × 𝐷))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 209  wa 400   = wceq 1563  wne 2960  wss 3907  wpss 3908  c0 4288   × cxp 5650
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-11 2194  ax-ext 2737  ax-sep 5251  ax-pr 5395
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1566  df-fal 1576  df-ex 1803  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-ral 3080  df-rex 3090  df-rab 3418  df-v 3459  df-dif 3910  df-un 3912  df-in 3914  df-ss 3924  df-pss 3927  df-nul 4289  df-if 4484  df-sn 4586  df-pr 4588  df-op 4592  df-br 5106  df-opab 5168  df-xp 5658  df-rel 5659  df-cnv 5660  df-dm 5662  df-rn 5663
This theorem is referenced by: (None)
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