Theorem nfcdeq 2971

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. (Contributed by Mario Carneiro, 11-Aug-2016.)

Hypotheses

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

Assertion

Ref	Expression
nfcdeq	⊢ (𝜑 ↔ 𝜓)

Distinct variable groups:   𝜓,𝑥   𝜑,𝑦

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

Proof of Theorem nfcdeq

Step	Hyp	Ref	Expression
1		nfcdeq.1	. . 3 ⊢ Ⅎ𝑥𝜑
2	1	sbf 1787	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑)
3		nfv 1538	. . 3 ⊢ Ⅎ𝑥𝜓
4		nfcdeq.2	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	4	cdeqri 2960	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
6	3, 5	sbie 1801	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
7	2, 6	bitr3i 186	1 ⊢ (𝜑 ↔ 𝜓)

Syntax hints:   ↔ wb 105  Ⅎwnf 1470  [wsb 1772  CondEqwcdeq 2957

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-5 1457  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544

This theorem depends on definitions:  df-bi 117  df-nf 1471  df-sb 1773  df-cdeq 2958

This theorem is referenced by:  nfccdeq  2972