Theorem cdeqi 3387

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

Hypothesis

Ref	Expression
cdeqi.1	⊢ (𝑥 = 𝑦 → 𝜑)

Assertion

Ref	Expression
cdeqi	⊢ CondEq(𝑥 = 𝑦 → 𝜑)

Proof of Theorem cdeqi

Step	Hyp	Ref	Expression
1		cdeqi.1	. 2 ⊢ (𝑥 = 𝑦 → 𝜑)
2		df-cdeq 3386	. 2 ⊢ (CondEq(𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜑))
3	1, 2	mpbir 220	1 ⊢ CondEq(𝑥 = 𝑦 → 𝜑)

Syntax hints:   → wi 4  CondEqwcdeq 3385

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 196  df-cdeq 3386

This theorem is referenced by:  cdeqth  3389  cdeqnot  3390  cdeqal  3391  cdeqab  3392  cdeqim  3395  cdeqcv  3396  cdeqeq  3397  cdeqel  3398  bj-cdeqab  31975