Theorem ss2abim 4007

Description: Class abstractions in a subclass relationship. Reverse direction of ss2ab 4008 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 2109	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑡 / 𝑥]𝜑 → [𝑡 / 𝑥]𝜓))
2		df-clab 2739	. . 3 ⊢ (𝑡 ∈ {𝑥 ∣ 𝜑} ↔ [𝑡 / 𝑥]𝜑)
3		df-clab 2739	. . 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 3936	1 ⊢ (∀𝑥(𝜑 → 𝜓) → {𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜓})

This proof depends on syntax axioms:   → wi 4  ∀wal 1568  [wsb 2099   ∈ wcel 2145  {cab 2738   ⊆ wss 3898

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

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2739  df-ss 3915

This theorem is used by:  ss2rabd  4019  moabex  5425