Theorem nfcdeq 3712

Description: If we have a conditional equality proof, where 𝜑 is 𝜑(𝑥) and 𝜓 is 𝜑(𝑦), and 𝜑(𝑥) in fact does not have 𝑥 free in it according to Ⅎ, then 𝜑(𝑥) ↔ 𝜑(𝑦) unconditionally. This proves that Ⅎ𝑥𝜑 is actually a not-free predicate. Usage of this theorem is discouraged because it depends on ax-13 2372. (Contributed by Mario Carneiro, 11-Aug-2016.) (New usage is discouraged.)

Hypotheses

Ref	Expression
nfcdeq.1	⊢ Ⅎ𝑥𝜑
nfcdeq.2	⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
nfcdeq	⊢ (𝜑 ↔ 𝜓)

Distinct variable group:   𝜓,𝑥

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

Proof of Theorem nfcdeq

Step	Hyp	Ref	Expression
1		nfcdeq.1	. . 3 ⊢ Ⅎ𝑥𝜑
2	1	sbf 2263	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑)
3		nfv 1917	. . 3 ⊢ Ⅎ𝑥𝜓
4		nfcdeq.2	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	4	cdeqri 3701	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
6	3, 5	sbie 2506	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
7	2, 6	bitr3i 276	1 ⊢ (𝜑 ↔ 𝜓)

Syntax hints:   ↔ wb 205  Ⅎwnf 1786  [wsb 2067  CondEqwcdeq 3698

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-10 2137  ax-12 2171  ax-13 2372

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-ex 1783  df-nf 1787  df-sb 2068  df-cdeq 3699

This theorem is referenced by:  nfccdeq  3713