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Theorem sbelx 2486
Description: Elimination of substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbelx (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))
Distinct variable groups:   𝑥,𝑦   𝜑,𝑥
Allowed substitution hint:   𝜑(𝑦)

Proof of Theorem sbelx
StepHypRef Expression
1 sbid2v 2485 . 2 ([𝑦 / 𝑥][𝑥 / 𝑦]𝜑𝜑)
2 sb5 2458 . 2 ([𝑦 / 𝑥][𝑥 / 𝑦]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))
31, 2bitr3i 266 1 (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))
Colors of variables: wff setvar class
Syntax hints:  wb 196  wa 383  wex 1744  [wsb 1937
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-10 2059  ax-12 2087  ax-13 2282
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-ex 1745  df-nf 1750  df-sb 1938
This theorem is referenced by:  pm13.196a  38932
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