Theorem disjss 39764

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 5849	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → ◡𝐴 ⊆ ◡𝐵)
2		funALTVss 39717	. . . 4 ⊢ (◡𝐴 ⊆ ◡𝐵 → ( FunALTV ◡𝐵 → FunALTV ◡𝐴))
3	1, 2	syl 18	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ( FunALTV ◡𝐵 → FunALTV ◡𝐴))
4		relss 5757	. . 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 39731	. 2 ⊢ ( Disj 𝐵 ↔ ( FunALTV ◡𝐵 ∧ Rel 𝐵))
7		dfdisjALTV 39731	. 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 3895  ^◡ccnv 5649  Rel wrel 5655   FunALTV wfunALTV 39149   Disj wdisjALTV 39152

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-11 2193  ax-ext 2732  ax-sep 5248  ax-pr 5390

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-br 5103  df-opab 5167  df-id 5545  df-xp 5656  df-rel 5657  df-cnv 5658  df-co 5659  df-dm 5660  df-rn 5661  df-res 5662  df-coss 39434  df-cnvrefrel 39540  df-funALTV 39700  df-disjALTV 39723

This theorem is used by:  disjssi  39765  disjssd  39766