Theorem ss2abim 3998

Description: Class abstractions in a subclass relationship. Reverse direction of ss2ab 3999 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 2083	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑡 / 𝑥]𝜑 → [𝑡 / 𝑥]𝜓))
2		df-clab 2719	. . 3 ⊢ (𝑡 ∈ {𝑥 ∣ 𝜑} ↔ [𝑡 / 𝑥]𝜑)
3		df-clab 2719	. . 3 ⊢ (𝑡 ∈ {𝑥 ∣ 𝜓} ↔ [𝑡 / 𝑥]𝜓)
4	1, 2, 3	3imtr4g 297	. 2 ⊢ (∀𝑥(𝜑 → 𝜓) → (𝑡 ∈ {𝑥 ∣ 𝜑} → 𝑡 ∈ {𝑥 ∣ 𝜓}))
5	4	ssrdv 3928	1 ⊢ (∀𝑥(𝜑 → 𝜓) → {𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜓})

Syntax hints:   → wi 4  ∀wal 1545  [wsb 2073   ∈ wcel 2119  {cab 2718   ⊆ wss 3890

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1787  df-sb 2074  df-clab 2719  df-ss 3907

This theorem is referenced by:  ss2rabd  4010  moabex  5404