Theorem nfcdeq 3734

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 2401. (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 2304	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑)
3		nfv 1946	. . 3 ⊢ Ⅎ𝑥𝜓
4		nfcdeq.2	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	4	cdeqri 3723	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
6	3, 5	sbie 2531	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
7	2, 6	bitr3i 280	1 ⊢ (𝜑 ↔ 𝜓)

This proof depends on syntax axioms:   ↔ wb 209  Ⅎwnf 1815  [wsb 2098  CondEqwcdeq 3720

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-10 2177  ax-12 2212  ax-13 2401

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1812  df-nf 1816  df-sb 2099  df-cdeq 3721

This theorem is used by:  nfccdeq  3735