Metamath Proof Explorer


Theorem sbt

Description: A substitution into a theorem yields a theorem. See sbtALT for a shorter proof requiring more axioms. See chvar and chvarv for versions using implicit substitution. (Contributed by NM, 21-Jan-2004) (Proof shortened by Andrew Salmon, 25-May-2011) (Proof shortened by Wolf Lammen, 20-Jul-2018) Revise df-sb . (Revised by Steven Nguyen, 6-Jul-2023)

Ref Expression
Hypothesis sbt.1 𝜑
Assertion sbt [ 𝑡 / 𝑥 ] 𝜑

Proof

Step Hyp Ref Expression
1 sbt.1 𝜑
2 1 a1i ( 𝑥 = 𝑦𝜑 )
3 2 ax-gen 𝑥 ( 𝑥 = 𝑦𝜑 )
4 3 a1i ( 𝑦 = 𝑡 → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )
5 4 ax-gen 𝑦 ( 𝑦 = 𝑡 → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )
6 df-sb ( [ 𝑡 / 𝑥 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑡 → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ) )
7 5 6 mpbir [ 𝑡 / 𝑥 ] 𝜑