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Theorem nfcdeq 3754
 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 2392. (Contributed by Mario Carneiro, 11-Aug-2016.) (New usage is discouraged.)
Hypotheses
Ref Expression
nfcdeq.1 𝑥𝜑
nfcdeq.2 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
nfcdeq (𝜑𝜓)
Distinct variable groups:   𝜓,𝑥   𝜑,𝑦
Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)

Proof of Theorem nfcdeq
StepHypRef Expression
1 nfcdeq.1 . . 3 𝑥𝜑
21sbf 2273 . 2 ([𝑦 / 𝑥]𝜑𝜑)
3 nfv 1916 . . 3 𝑥𝜓
4 nfcdeq.2 . . . 4 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
54cdeqri 3743 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
63, 5sbie 2546 . 2 ([𝑦 / 𝑥]𝜑𝜓)
72, 6bitr3i 280 1 (𝜑𝜓)
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 209  Ⅎwnf 1785  [wsb 2070  CondEqwcdeq 3740 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-10 2146  ax-12 2179  ax-13 2392 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-nf 1786  df-sb 2071  df-cdeq 3741 This theorem is referenced by:  nfccdeq  3755
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