Theorem disjss1 4112

Description: A subset of a disjoint collection is disjoint. (Contributed by Mario Carneiro, 14-Nov-2016.)

Assertion

Ref	Expression
disjss1	⊢ (𝐴 ⊆ 𝐵 → (Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝐶(𝑥)

Proof of Theorem disjss1

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ssel 3242	. . . . . 6 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	anim1d 336	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶) → (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))
3	2	alrimiv 1927	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶) → (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))
4		moim 2151	. . . 4 ⊢ (∀𝑥((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶) → (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)) → (∃*𝑥(𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶) → ∃*𝑥(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)))
5	3, 4	syl 14	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (∃*𝑥(𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶) → ∃*𝑥(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)))
6	5	alimdv 1932	. 2 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑦∃*𝑥(𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶) → ∀𝑦∃*𝑥(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)))
7		dfdisj2 4108	. 2 ⊢ (Disj 𝑥 ∈ 𝐵 𝐶 ↔ ∀𝑦∃*𝑥(𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))
8		dfdisj2 4108	. 2 ⊢ (Disj 𝑥 ∈ 𝐴 𝐶 ↔ ∀𝑦∃*𝑥(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶))
9	6, 7, 8	3imtr4g 205	1 ⊢ (𝐴 ⊆ 𝐵 → (Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104  ∀wal 1400  ∃*wmo 2087   ∈ wcel 2209   ⊆ wss 3220  Disj wdisj 4106

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-rmo 2536  df-in 3226  df-ss 3233  df-disj 4107

This theorem is used by:  disjeq1  4113  disjx0  4129  fsumiun  12263