Theorem elscottrankss 35633

Description: Relationship between the ranks of an element in a Scott's trick set and an element in the input set. (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 35627	. . 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 6879	. . . 4 ⊢ (𝑥 = 𝐶 → (rank‘𝑥) = (rank‘𝐶))
4	3	sseq2d 3963	. . 3 ⊢ (𝑥 = 𝐶 → ((rank‘𝐴) ⊆ (rank‘𝑥) ↔ (rank‘𝐴) ⊆ (rank‘𝐶)))
5	4	rspccva 3575	. 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 1570   ∈ wcel 2145  ∀wral 3076   ⊆ wss 3899  ‘cfv 6533  rankcrnk 9748  Scott cscott 9870

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-ext 2732

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-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-iota 6489  df-fv 6541  df-scott 9871

This theorem is used by:  nelscottrankgt  35635  rankscottu  35639