Theorem ssxp2 4758

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 4752	. . . . . 6 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ran (𝐶 × 𝐴) = 𝐴)
2	1	adantr 261	. . . . 5 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → ran (𝐶 × 𝐴) = 𝐴)
3		rnss 4564	. . . . . 6 ⊢ ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) → ran (𝐶 × 𝐴) ⊆ ran (𝐶 × 𝐵))
4	3	adantl 262	. . . . 5 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → ran (𝐶 × 𝐴) ⊆ ran (𝐶 × 𝐵))
5	2, 4	eqsstr3d 2980	. . . 4 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → 𝐴 ⊆ ran (𝐶 × 𝐵))
6		rnxpss 4754	. . . 4 ⊢ ran (𝐶 × 𝐵) ⊆ 𝐵
7	5, 6	syl6ss 2957	. . 3 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → 𝐴 ⊆ 𝐵)
8	7	ex 108	. 2 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) → 𝐴 ⊆ 𝐵))
9		xpss2 4449	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵))
10	8, 9	impbid1 130	1 ⊢ (∃𝑥 𝑥 ∈ 𝐶 → ((𝐶 × 𝐴) ⊆ (𝐶 × 𝐵) ↔ 𝐴 ⊆ 𝐵))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1243  ∃wex 1381   ∈ wcel 1393   ⊆ wss 2917   × cxp 4343  ran crn 4346

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-14 1405  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022  ax-sep 3875  ax-pow 3927  ax-pr 3944

This theorem depends on definitions:  df-bi 110  df-3an 887  df-tru 1246  df-nf 1350  df-sb 1646  df-eu 1903  df-mo 1904  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311  df-rex 2312  df-v 2559  df-un 2922  df-in 2924  df-ss 2931  df-pw 3361  df-sn 3381  df-pr 3382  df-op 3384  df-br 3765  df-opab 3819  df-xp 4351  df-rel 4352  df-cnv 4353  df-dm 4355  df-rn 4356

This theorem is referenced by:  xpcanm  4760