Theorem cdeqab 2941

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 2937	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	abbidv 2284	. 2 ⊢ (𝑥 = 𝑦 → {𝑧 ∣ 𝜑} = {𝑧 ∣ 𝜓})
4	3	cdeqi 2936	1 ⊢ CondEq(𝑥 = 𝑦 → {𝑧 ∣ 𝜑} = {𝑧 ∣ 𝜓})

Syntax hints:   ↔ wb 104   = wceq 1343  {cab 2151  CondEqwcdeq 2934

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-11 1494  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-cdeq 2935

This theorem is referenced by: (None)