Theorem cdeqab1 3020

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
cdeqab1	⊢ CondEq(𝑥 = 𝑦 → {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓})

Distinct variable groups:   𝜓,𝑥   𝜑,𝑦

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

Proof of Theorem cdeqab1

Step	Hyp	Ref	Expression
1		cdeqnot.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	cdeqri 3014	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	cbvabv 2354	. 2 ⊢ {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓}
4	3	cdeqth 3015	1 ⊢ CondEq(𝑥 = 𝑦 → {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓})

Syntax hints:   ↔ wb 105   = wceq 1395  {cab 2215  CondEqwcdeq 3011

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-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-cdeq 3012

This theorem is referenced by: (None)