Theorem ssxp1 5175

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 4954	. . . . . 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 4932	. . . . . 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 3263	. . . 4 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐴 × 𝐶) ⊆ (𝐵 × 𝐶)) → 𝐴 ⊆ dom (𝐵 × 𝐶))
6		dmxpss 5169	. . . 4 ⊢ dom (𝐵 × 𝐶) ⊆ 𝐵
7	5, 6	sstrdi 3238	. . 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 4838	. 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 1397  ∃wex 1540   ∈ wcel 2201   ⊆ wss 3199   × cxp 4725  dom cdm 4727

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-14 2204  ax-ext 2212  ax-sep 4208  ax-pow 4266  ax-pr 4301

This theorem depends on definitions:  df-bi 117  df-3an 1006  df-tru 1400  df-nf 1509  df-sb 1810  df-eu 2081  df-mo 2082  df-clab 2217  df-cleq 2223  df-clel 2226  df-nfc 2362  df-ral 2514  df-rex 2515  df-v 2803  df-un 3203  df-in 3205  df-ss 3212  df-pw 3655  df-sn 3676  df-pr 3677  df-op 3679  df-br 4090  df-opab 4152  df-xp 4733  df-dm 4737

This theorem is referenced by:  xpcan2m  5179