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Theorem xppss12 42812
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 4051 . . 3 (𝐴𝐵𝐴𝐵)
2 pssss 4051 . . 3 (𝐶𝐷𝐶𝐷)
3 xpss12 5660 . . 3 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷))
41, 2, 3syl2an 605 . 2 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷))
5 simpl 486 . . . . 5 ((𝐴𝐵𝐶𝐷) → 𝐴𝐵)
6 pssne 4052 . . . . . 6 (𝐴𝐵𝐴𝐵)
76necomd 3011 . . . . 5 (𝐴𝐵𝐵𝐴)
8 neneq 2962 . . . . . 6 (𝐵𝐴 → ¬ 𝐵 = 𝐴)
98intnanrd 493 . . . . 5 (𝐵𝐴 → ¬ (𝐵 = 𝐴𝐷 = 𝐶))
105, 7, 93syl 18 . . . 4 ((𝐴𝐵𝐶𝐷) → ¬ (𝐵 = 𝐴𝐷 = 𝐶))
11 pssn0 42810 . . . . 5 (𝐴𝐵𝐵 ≠ ∅)
12 pssn0 42810 . . . . 5 (𝐶𝐷𝐷 ≠ ∅)
13 xp11 6157 . . . . 5 ((𝐵 ≠ ∅ ∧ 𝐷 ≠ ∅) → ((𝐵 × 𝐷) = (𝐴 × 𝐶) ↔ (𝐵 = 𝐴𝐷 = 𝐶)))
1411, 12, 13syl2an 605 . . . 4 ((𝐴𝐵𝐶𝐷) → ((𝐵 × 𝐷) = (𝐴 × 𝐶) ↔ (𝐵 = 𝐴𝐷 = 𝐶)))
1510, 14mtbird 327 . . 3 ((𝐴𝐵𝐶𝐷) → ¬ (𝐵 × 𝐷) = (𝐴 × 𝐶))
16 neqne 2964 . . . 4 (¬ (𝐵 × 𝐷) = (𝐴 × 𝐶) → (𝐵 × 𝐷) ≠ (𝐴 × 𝐶))
1716necomd 3011 . . 3 (¬ (𝐵 × 𝐷) = (𝐴 × 𝐶) → (𝐴 × 𝐶) ≠ (𝐵 × 𝐷))
1815, 17syl 17 . 2 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ≠ (𝐵 × 𝐷))
19 df-pss 3924 . 2 ((𝐴 × 𝐶) ⊊ (𝐵 × 𝐷) ↔ ((𝐴 × 𝐶) ⊆ (𝐵 × 𝐷) ∧ (𝐴 × 𝐶) ≠ (𝐵 × 𝐷)))
204, 18, 19sylanbrc 592 1 ((𝐴𝐵𝐶𝐷) → (𝐴 × 𝐶) ⊊ (𝐵 × 𝐷))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 208  wa 399   = wceq 1559  wne 2956  wss 3904  wpss 3905  c0 4285   × cxp 5643
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-11 2190  ax-ext 2733  ax-sep 5245  ax-pr 5389
This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1099  df-tru 1562  df-fal 1572  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-ral 3076  df-rex 3086  df-rab 3414  df-v 3455  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-pss 3924  df-nul 4286  df-if 4480  df-sn 4582  df-pr 4584  df-op 4588  df-br 5100  df-opab 5162  df-xp 5651  df-rel 5652  df-cnv 5653  df-dm 5655  df-rn 5656
This theorem is referenced by: (None)
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