Theorem ssxp1 5165

Description: Cross product subset cancellation. (Contributed by Jim Kingdon, 14-Dec-2018.)

Assertion

Ref	Expression
ssxp1	⊢ (∃𝑥 𝑥 ∈ 𝐶 → ((𝐴 × 𝐶) ⊆ (𝐵 × 𝐶) ↔ 𝐴 ⊆ 𝐵))

Distinct variable group:   𝑥,𝐶

Allowed substitution hints:   𝐴(𝑥)   𝐵(𝑥)

Proof of Theorem ssxp1

Step	Hyp	Ref	Expression
1		dmxpm 4944	. . . . . 6 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → dom (𝐴 × 𝐶) = 𝐴)
2	1	adantr 276	. . . . 5 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐴 × 𝐶) ⊆ (𝐵 × 𝐶)) → dom (𝐴 × 𝐶) = 𝐴)
3		dmss 4922	. . . . . 6 ⊢ ((𝐴 × 𝐶) ⊆ (𝐵 × 𝐶) → dom (𝐴 × 𝐶) ⊆ dom (𝐵 × 𝐶))
4	3	adantl 277	. . . . 5 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐴 × 𝐶) ⊆ (𝐵 × 𝐶)) → dom (𝐴 × 𝐶) ⊆ dom (𝐵 × 𝐶))
5	2, 4	eqsstrrd 3261	. . . 4 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐴 × 𝐶) ⊆ (𝐵 × 𝐶)) → 𝐴 ⊆ dom (𝐵 × 𝐶))
6		dmxpss 5159	. . . 4 ⊢ dom (𝐵 × 𝐶) ⊆ 𝐵
7	5, 6	sstrdi 3236	. . 3 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐴 × 𝐶) ⊆ (𝐵 × 𝐶)) → 𝐴 ⊆ 𝐵)
8	7	ex 115	. 2 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ((𝐴 × 𝐶) ⊆ (𝐵 × 𝐶) → 𝐴 ⊆ 𝐵))
9		xpss1 4829	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐶))
10	8, 9	impbid1 142	1 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ((𝐴 × 𝐶) ⊆ (𝐵 × 𝐶) ↔ 𝐴 ⊆ 𝐵))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1395  ∃wex 1538   ∈ wcel 2200   ⊆ wss 3197   × cxp 4717  dom cdm 4719

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-14 2203  ax-ext 2211  ax-sep 4202  ax-pow 4258  ax-pr 4293

This theorem depends on definitions:  df-bi 117  df-3an 1004  df-tru 1398  df-nf 1507  df-sb 1809  df-eu 2080  df-mo 2081  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-rex 2514  df-v 2801  df-un 3201  df-in 3203  df-ss 3210  df-pw 3651  df-sn 3672  df-pr 3673  df-op 3675  df-br 4084  df-opab 4146  df-xp 4725  df-dm 4729

This theorem is referenced by:  xpcan2m  5169