Theorem ss2abim 4008

Description: Class abstractions in a subclass relationship. Reverse direction of ss2ab 4009 which requires fewer axioms. (Contributed by SN, 22-Dec-2024.)

Assertion

Ref	Expression
ss2abim	⊢ (∀𝑥(𝜑 → 𝜓) → {𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜓})

Proof of Theorem ss2abim

Dummy variable 𝑡 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		spsbim 2099	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑡 / 𝑥]𝜑 → [𝑡 / 𝑥]𝜓))
2		df-clab 2735	. . 3 ⊢ (𝑡 ∈ {𝑥 ∣ 𝜑} ↔ [𝑡 / 𝑥]𝜑)
3		df-clab 2735	. . 3 ⊢ (𝑡 ∈ {𝑥 ∣ 𝜓} ↔ [𝑡 / 𝑥]𝜓)
4	1, 2, 3	3imtr4g 298	. 2 ⊢ (∀𝑥(𝜑 → 𝜓) → (𝑡 ∈ {𝑥 ∣ 𝜑} → 𝑡 ∈ {𝑥 ∣ 𝜓}))
5	4	ssrdv 3937	1 ⊢ (∀𝑥(𝜑 → 𝜓) → {𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜓})

Syntax hints:   → wi 4  ∀wal 1552  [wsb 2084   ∈ wcel 2136  {cab 2734   ⊆ wss 3899

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1809  ax-4 1823  ax-5 1924

This theorem depends on definitions:  df-bi 209  df-an 399  df-sb 2085  df-clab 2735  df-ss 3916

This theorem is referenced by:  ss2rabd  4020  moabex  5419