Theorem elscottrankss 35576

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 35570	. . 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 6885	. . . 4 ⊢ (𝑥 = 𝐶 → (rank‘𝑥) = (rank‘𝐶))
4	3	sseq2d 3970	. . 3 ⊢ (𝑥 = 𝐶 → ((rank‘𝐴) ⊆ (rank‘𝑥) ↔ (rank‘𝐴) ⊆ (rank‘𝐶)))
5	4	rspccva 3582	. 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 2146  ∀wral 3081   ⊆ wss 3906  ‘cfv 6540  rankcrnk 9742  Scott cscott 9864

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 2148  ax-9 2156  ax-ext 2737

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 2744  df-cleq 2757  df-clel 2840  df-ral 3082  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-uni 4875  df-br 5112  df-iota 6496  df-fv 6548  df-scott 9865

This theorem is used by:  nelscottrankgt  35578  rankscottu  35582