Theorem cdeqab 2975

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 2971	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	abbidv 2311	. 2 ⊢ (𝑥 = 𝑦 → {𝑧 ∣ 𝜑} = {𝑧 ∣ 𝜓})
4	3	cdeqi 2970	1 ⊢ CondEq(𝑥 = 𝑦 → {𝑧 ∣ 𝜑} = {𝑧 ∣ 𝜓})

Syntax hints:   ↔ wb 105   = wceq 1364  {cab 2179  CondEqwcdeq 2968

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-11 1517  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-cdeq 2969

This theorem is referenced by: (None)