Theorem disjss 39579

Description: Subclass theorem for disjoints. (Contributed by Peter Mazsa, 28-Oct-2020.) (Revised by Peter Mazsa, 22-Sep-2021.)

Assertion

Ref	Expression
disjss	⊢ (𝐴 ⊆ 𝐵 → ( Disj 𝐵 → Disj 𝐴))

Proof of Theorem disjss

Step	Hyp	Ref	Expression
1		cnvss 5852	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → ◡𝐴 ⊆ ◡𝐵)
2		funALTVss 39532	. . . 4 ⊢ (◡𝐴 ⊆ ◡𝐵 → ( FunALTV ◡𝐵 → FunALTV ◡𝐴))
3	1, 2	syl 18	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ( FunALTV ◡𝐵 → FunALTV ◡𝐴))
4		relss 5762	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (Rel 𝐵 → Rel 𝐴))
5	3, 4	anim12d 621	. 2 ⊢ (𝐴 ⊆ 𝐵 → (( FunALTV ◡𝐵 ∧ Rel 𝐵) → ( FunALTV ◡𝐴 ∧ Rel 𝐴)))
6		dfdisjALTV 39546	. 2 ⊢ ( Disj 𝐵 ↔ ( FunALTV ◡𝐵 ∧ Rel 𝐵))
7		dfdisjALTV 39546	. 2 ⊢ ( Disj 𝐴 ↔ ( FunALTV ◡𝐴 ∧ Rel 𝐴))
8	5, 6, 7	3imtr4g 299	1 ⊢ (𝐴 ⊆ 𝐵 → ( Disj 𝐵 → Disj 𝐴))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ⊆ wss 3899  ^◡ccnv 5654  Rel wrel 5660   FunALTV wfunALTV 38964   Disj wdisjALTV 38967

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-11 2194  ax-ext 2732  ax-sep 5251  ax-pr 5398

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-id 5550  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-coss 39249  df-cnvrefrel 39355  df-funALTV 39515  df-disjALTV 39538

This theorem is used by:  disjssi  39580  disjssd  39581