Theorem cdeqab 3732

Description: Distribute conditional equality over abstraction. (Contributed by Mario Carneiro, 11-Aug-2016.)

Hypothesis

Ref	Expression
cdeqnot.1	⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
cdeqab	⊢ CondEq(𝑥 = 𝑦 → {𝑧 ∣ 𝜑} = {𝑧 ∣ 𝜓})

Distinct variable groups:   𝑥,𝑧   𝑦,𝑧

Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)   𝜓(𝑥,𝑦,𝑧)

Proof of Theorem cdeqab

Step	Hyp	Ref	Expression
1		cdeqnot.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	cdeqri 3728	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	abbidv 2827	. 2 ⊢ (𝑥 = 𝑦 → {𝑧 ∣ 𝜑} = {𝑧 ∣ 𝜓})
4	3	cdeqi 3727	1 ⊢ CondEq(𝑥 = 𝑦 → {𝑧 ∣ 𝜑} = {𝑧 ∣ 𝜓})

Syntax hints:   ↔ wb 208   = wceq 1559  {cab 2739  CondEqwcdeq 3725

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-cdeq 3726

This theorem is referenced by: (None)