Theorem cdeqab 3727

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 3723	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	abbidv 2826	. 2 ⊢ (𝑥 = 𝑦 → {𝑧 ∣ 𝜑} = {𝑧 ∣ 𝜓})
4	3	cdeqi 3722	1 ⊢ CondEq(𝑥 = 𝑦 → {𝑧 ∣ 𝜑} = {𝑧 ∣ 𝜓})

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1569  {cab 2738  CondEqwcdeq 3720

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-cdeq 3721

This theorem is used by: (None)