Metamath Proof Explorer


Theorem sbiedvw

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbievw ). Version of sbied and sbiedv with more disjoint variable conditions, requiring fewer axioms. (Contributed by NM, 30-Jun-1994) (Revised by GG, 29-Jan-2024)

Ref Expression
Hypothesis sbiedvw.1 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝜓 ↔ 𝜒 ) )
Assertion sbiedvw ( 𝜑 → ( [ 𝑦 / 𝑥 ] 𝜓 ↔ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 sbiedvw.1 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝜓 ↔ 𝜒 ) )
2 sbrimvw ⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )
3 1 expcom ⊢ ( 𝑥 = 𝑦 → ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) )
4 3 pm5.74d ⊢ ( 𝑥 = 𝑦 → ( ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → 𝜒 ) ) )
5 4 sbievw ⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → 𝜒 ) )
6 2 5 bitr3i ⊢ ( ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) ↔ ( 𝜑 → 𝜒 ) )
7 6 pm5.74ri ⊢ ( 𝜑 → ( [ 𝑦 / 𝑥 ] 𝜓 ↔ 𝜒 ) )