Theorem ssxp2 5205

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

Assertion

Ref	Expression
ssxp2	⊢ (∃𝑥 𝑥 ∈ 𝐶 → ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) ↔ 𝐴 ⊆ 𝐵))

Distinct variable group:   𝑥,𝐶

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

Proof of Theorem ssxp2

Step	Hyp	Ref	Expression
1		rnxpm 5197	. . . . . 6 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ran (𝐶 × 𝐴) = 𝐴)
2	1	adantr 276	. . . . 5 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → ran (𝐶 × 𝐴) = 𝐴)
3		rnss 4992	. . . . . 6 ⊢ ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) → ran (𝐶 × 𝐴) ⊆ ran (𝐶 × 𝐵))
4	3	adantl 277	. . . . 5 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → ran (𝐶 × 𝐴) ⊆ ran (𝐶 × 𝐵))
5	2, 4	eqsstrrd 3279	. . . 4 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → 𝐴 ⊆ ran (𝐶 × 𝐵))
6		rnxpss 5199	. . . 4 ⊢ ran (𝐶 × 𝐵) ⊆ 𝐵
7	5, 6	sstrdi 3254	. . 3 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → 𝐴 ⊆ 𝐵)
8	7	ex 115	. 2 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) → 𝐴 ⊆ 𝐵))
9		xpss2 4866	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵))
10	8, 9	impbid1 142	1 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) ↔ 𝐴 ⊆ 𝐵))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1398  ∃wex 1541   ∈ wcel 2205   ⊆ wss 3214   × cxp 4752  ran crn 4755

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 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-14 2208  ax-ext 2216  ax-sep 4233  ax-pow 4292  ax-pr 4327

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1812  df-eu 2085  df-mo 2086  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-ral 2527  df-rex 2528  df-v 2817  df-un 3218  df-in 3220  df-ss 3227  df-pw 3676  df-sn 3700  df-pr 3701  df-op 3703  df-br 4115  df-opab 4177  df-xp 4760  df-rel 4761  df-cnv 4762  df-dm 4764  df-rn 4765

This theorem is referenced by:  xpcanm  5207