Theorem ssxp2 5125

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 5117	. . . . . 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 4913	. . . . . 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 3231	. . . 4 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → 𝐴 ⊆ ran (𝐶 × 𝐵))
6		rnxpss 5119	. . . 4 ⊢ ran (𝐶 × 𝐵) ⊆ 𝐵
7	5, 6	sstrdi 3206	. . 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 4790	. 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 1373  ∃wex 1516   ∈ wcel 2177   ⊆ wss 3167   × cxp 4677  ran crn 4680

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-14 2180  ax-ext 2188  ax-sep 4166  ax-pow 4222  ax-pr 4257

This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-nf 1485  df-sb 1787  df-eu 2058  df-mo 2059  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-ral 2490  df-rex 2491  df-v 2775  df-un 3171  df-in 3173  df-ss 3180  df-pw 3619  df-sn 3640  df-pr 3641  df-op 3643  df-br 4048  df-opab 4110  df-xp 4685  df-rel 4686  df-cnv 4687  df-dm 4689  df-rn 4690

This theorem is referenced by:  xpcanm  5127