Theorem xpidtr 6115

Description: A Cartesian square is a transitive relation. (Contributed by FL, 31-Jul-2009.)

Assertion

Ref	Expression
xpidtr	⊢ ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) ⊆ (𝐴 × 𝐴)

Proof of Theorem xpidtr

Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		brxp 5699	. . . . . 6 ⊢ (𝑥(𝐴 × 𝐴)𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐴))
2		brxp 5699	. . . . . . . . 9 ⊢ (𝑦(𝐴 × 𝐴)𝑧 ↔ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴))
3		brxp 5699	. . . . . . . . . 10 ⊢ (𝑥(𝐴 × 𝐴)𝑧 ↔ (𝑥 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴))
4	3	simplbi2com 508	. . . . . . . . 9 ⊢ (𝑧 ∈ 𝐴 → (𝑥 ∈ 𝐴 → 𝑥(𝐴 × 𝐴)𝑧))
5	2, 4	simplbiim 514	. . . . . . . 8 ⊢ (𝑦(𝐴 × 𝐴)𝑧 → (𝑥 ∈ 𝐴 → 𝑥(𝐴 × 𝐴)𝑧))
6	5	com12 33	. . . . . . 7 ⊢ (𝑥 ∈ 𝐴 → (𝑦(𝐴 × 𝐴)𝑧 → 𝑥(𝐴 × 𝐴)𝑧))
7	6	adantr 486	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐴) → (𝑦(𝐴 × 𝐴)𝑧 → 𝑥(𝐴 × 𝐴)𝑧))
8	1, 7	sylbi 220	. . . . 5 ⊢ (𝑥(𝐴 × 𝐴)𝑦 → (𝑦(𝐴 × 𝐴)𝑧 → 𝑥(𝐴 × 𝐴)𝑧))
9	8	imp 412	. . . 4 ⊢ ((𝑥(𝐴 × 𝐴)𝑦 ∧ 𝑦(𝐴 × 𝐴)𝑧) → 𝑥(𝐴 × 𝐴)𝑧)
10	9	ax-gen 1827	. . 3 ⊢ ∀𝑧((𝑥(𝐴 × 𝐴)𝑦 ∧ 𝑦(𝐴 × 𝐴)𝑧) → 𝑥(𝐴 × 𝐴)𝑧)
11	10	gen2 1828	. 2 ⊢ ∀𝑥∀𝑦∀𝑧((𝑥(𝐴 × 𝐴)𝑦 ∧ 𝑦(𝐴 × 𝐴)𝑧) → 𝑥(𝐴 × 𝐴)𝑧)
12		cotr 6105	. 2 ⊢ (((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) ⊆ (𝐴 × 𝐴) ↔ ∀𝑥∀𝑦∀𝑧((𝑥(𝐴 × 𝐴)𝑦 ∧ 𝑦(𝐴 × 𝐴)𝑧) → 𝑥(𝐴 × 𝐴)𝑧))
13	11, 12	mpbir 234	1 ⊢ ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) ⊆ (𝐴 × 𝐴)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401  ∀wal 1567   ∈ wcel 2144   ⊆ wss 3895   class class class wbr 5102   × cxp 5648   ∘ ccom 5654

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-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-xp 5656  df-rel 5657  df-co 5659

This theorem is used by:  trinxp  6118  xpider  8809  trust  24548  rtrclex  44659  rtrclexi  44663