Theorem ssxp2 5048

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 5040	. . . . . 6 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ran (𝐶 × 𝐴) = 𝐴)
2	1	adantr 274	. . . . 5 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → ran (𝐶 × 𝐴) = 𝐴)
3		rnss 4841	. . . . . 6 ⊢ ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) → ran (𝐶 × 𝐴) ⊆ ran (𝐶 × 𝐵))
4	3	adantl 275	. . . . 5 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → ran (𝐶 × 𝐴) ⊆ ran (𝐶 × 𝐵))
5	2, 4	eqsstrrd 3184	. . . 4 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → 𝐴 ⊆ ran (𝐶 × 𝐵))
6		rnxpss 5042	. . . 4 ⊢ ran (𝐶 × 𝐵) ⊆ 𝐵
7	5, 6	sstrdi 3159	. . 3 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → 𝐴 ⊆ 𝐵)
8	7	ex 114	. 2 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) → 𝐴 ⊆ 𝐵))
9		xpss2 4722	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵))
10	8, 9	impbid1 141	1 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) ↔ 𝐴 ⊆ 𝐵))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104   = wceq 1348  ∃wex 1485   ∈ wcel 2141   ⊆ wss 3121   × cxp 4609  ran crn 4612

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-14 2144  ax-ext 2152  ax-sep 4107  ax-pow 4160  ax-pr 4194

This theorem depends on definitions:  df-bi 116  df-3an 975  df-tru 1351  df-nf 1454  df-sb 1756  df-eu 2022  df-mo 2023  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rex 2454  df-v 2732  df-un 3125  df-in 3127  df-ss 3134  df-pw 3568  df-sn 3589  df-pr 3590  df-op 3592  df-br 3990  df-opab 4051  df-xp 4617  df-rel 4618  df-cnv 4619  df-dm 4621  df-rn 4622

This theorem is referenced by:  xpcanm  5050