Theorem ss2abim 4008

Description: Class abstractions in a subclass relationship. Reverse direction of ss2ab 4009 which requires fewer axioms. (Contributed by SN, 2-Feb-2026.)

Assertion

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

Proof of Theorem ss2abim

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		spsbim 2075	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
2	1	alrimiv 1928	. 2 ⊢ (∀𝑥(𝜑 → 𝜓) → ∀𝑦([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
3		df-ss 3914	. . 3 ⊢ ({𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} → 𝑦 ∈ {𝑥 ∣ 𝜓}))
4		df-clab 2710	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
5		df-clab 2710	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
6	4, 5	imbi12i 350	. . . 4 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} → 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
7	6	albii 1820	. . 3 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} → 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑦([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
8	3, 7	bitr2i 276	. 2 ⊢ (∀𝑦([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ↔ {𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜓})
9	2, 8	sylib 218	1 ⊢ (∀𝑥(𝜑 → 𝜓) → {𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜓})

Syntax hints:   → wi 4  ∀wal 1539  [wsb 2067   ∈ wcel 2111  {cab 2709   ⊆ wss 3897

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911

This theorem depends on definitions:  df-bi 207  df-sb 2068  df-clab 2710  df-ss 3914

This theorem is referenced by:  ss2rabd  4020  moabex  5401