Theorem ssxp1 5222

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 5000	. . . . . 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 4978	. . . . . 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 3285	. . . 4 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐴 × 𝐶) ⊆ (𝐵 × 𝐶)) → 𝐴 ⊆ dom (𝐵 × 𝐶))
6		dmxpss 5216	. . . 4 ⊢ dom (𝐵 × 𝐶) ⊆ 𝐵
7	5, 6	sstrdi 3260	. . 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 4883	. 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 1402  ∃wex 1545   ∈ wcel 2209   ⊆ wss 3220   × cxp 4770  dom cdm 4772

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 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-sep 4247  ax-pow 4309  ax-pr 4344

This theorem depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823  df-un 3224  df-in 3226  df-ss 3233  df-pw 3690  df-sn 3714  df-pr 3715  df-op 3717  df-br 4129  df-opab 4191  df-xp 4778  df-dm 4782

This theorem is referenced by:  xpcan2m  5226