Theorem ss2abim 4013

Description: Class abstractions in a subclass relationship. Reverse direction of ss2ab 4014 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 2105	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑡 / 𝑥]𝜑 → [𝑡 / 𝑥]𝜓))
2		df-clab 2741	. . 3 ⊢ (𝑡 ∈ {𝑥 ∣ 𝜑} ↔ [𝑡 / 𝑥]𝜑)
3		df-clab 2741	. . 3 ⊢ (𝑡 ∈ {𝑥 ∣ 𝜓} ↔ [𝑡 / 𝑥]𝜓)
4	1, 2, 3	3imtr4g 299	. 2 ⊢ (∀𝑥(𝜑 → 𝜓) → (𝑡 ∈ {𝑥 ∣ 𝜑} → 𝑡 ∈ {𝑥 ∣ 𝜓}))
5	4	ssrdv 3942	1 ⊢ (∀𝑥(𝜑 → 𝜓) → {𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜓})

This proof depends on syntax axioms:   → wi 4  ∀wal 1567  [wsb 2095   ∈ wcel 2142  {cab 2740   ⊆ wss 3904

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037

This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1809  df-sb 2096  df-clab 2741  df-ss 3921

This theorem is used by:  ss2rabd  4025  moabex  5438