Theorem ssxp2 5064

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 5056	. . . . . 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 4855	. . . . . 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 3192	. . . 4 ⊢ ((∃𝑥 𝑥 ∈ 𝐶 ∧ (𝐶 × 𝐴) ⊆ (𝐶 × 𝐵)) → 𝐴 ⊆ ran (𝐶 × 𝐵))
6		rnxpss 5058	. . . 4 ⊢ ran (𝐶 × 𝐵) ⊆ 𝐵
7	5, 6	sstrdi 3167	. . 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 4736	. 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 1353  ∃wex 1492   ∈ wcel 2148   ⊆ wss 3129   × cxp 4623  ran crn 4626

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-14 2151  ax-ext 2159  ax-sep 4120  ax-pow 4173  ax-pr 4208

This theorem depends on definitions:  df-bi 117  df-3an 980  df-tru 1356  df-nf 1461  df-sb 1763  df-eu 2029  df-mo 2030  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-rex 2461  df-v 2739  df-un 3133  df-in 3135  df-ss 3142  df-pw 3577  df-sn 3598  df-pr 3599  df-op 3601  df-br 4003  df-opab 4064  df-xp 4631  df-rel 4632  df-cnv 4633  df-dm 4635  df-rn 4636

This theorem is referenced by:  xpcanm  5066