Theorem elscottrankss 35813

Description: Relationship between the ranks of an element in a Scott's trick set and an element in the input class. (Contributed by BTernaryTau, 3-Jul-2026.)

Assertion

Ref	Expression
elscottrankss	⊢ ((𝐴 ∈ Scott 𝐵 ∧ 𝐶 ∈ 𝐵) → (rank‘𝐴) ⊆ (rank‘𝐶))

Proof of Theorem elscottrankss

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		elscott 35807	. . 3 ⊢ (𝐴 ∈ Scott 𝐵 ↔ (𝐴 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐵 (rank‘𝐴) ⊆ (rank‘𝑥)))
2	1	simprbi 503	. 2 ⊢ (𝐴 ∈ Scott 𝐵 → ∀𝑥 ∈ 𝐵 (rank‘𝐴) ⊆ (rank‘𝑥))
3		fveq2 6896	. . . 4 ⊢ (𝑥 = 𝐶 → (rank‘𝑥) = (rank‘𝐶))
4	3	sseq2d 3945	. . 3 ⊢ (𝑥 = 𝐶 → ((rank‘𝐴) ⊆ (rank‘𝑥) ↔ (rank‘𝐴) ⊆ (rank‘𝐶)))
5	4	rspccva 3579	. 2 ⊢ ((∀𝑥 ∈ 𝐵 (rank‘𝐴) ⊆ (rank‘𝑥) ∧ 𝐶 ∈ 𝐵) → (rank‘𝐴) ⊆ (rank‘𝐶))
6	2, 5	sylan 592	1 ⊢ ((𝐴 ∈ Scott 𝐵 ∧ 𝐶 ∈ 𝐵) → (rank‘𝐴) ⊆ (rank‘𝐶))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1569   ∈ wcel 2146  ∀wral 3080   ⊆ wss 3900  ‘cfv 6547  rankcrnk 9780  Scott cscott 9941

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 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

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 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-br 5107  df-iota 6501  df-fv 6555  df-scott 9942

This theorem is used by:  nelscottrankgt  35815  rankscottu  35821