Theorem sbcthdv 3000

Description: Deduction version of sbcth 2999. (Contributed by NM, 30-Nov-2005.) (Proof shortened by Andrew Salmon, 8-Jun-2011.)

Hypothesis

Ref	Expression
sbcthdv.1	⊢ (𝜑 → 𝜓)

Assertion

Ref	Expression
sbcthdv	⊢ ((𝜑 ∧ 𝐴 ∈ 𝑉) → [𝐴 / 𝑥]𝜓)

Distinct variable group:   𝜑,𝑥

Allowed substitution hints:   𝜓(𝑥)   𝐴(𝑥)   𝑉(𝑥)

Proof of Theorem sbcthdv

Step	Hyp	Ref	Expression
1		sbcthdv.1	. . 3 ⊢ (𝜑 → 𝜓)
2	1	alrimiv 1885	. 2 ⊢ (𝜑 → ∀𝑥𝜓)
3		spsbc 2997	. 2 ⊢ (𝐴 ∈ 𝑉 → (∀𝑥𝜓 → [𝐴 / 𝑥]𝜓))
4	2, 3	mpan9 281	1 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑉) → [𝐴 / 𝑥]𝜓)

Syntax hints:   → wi 4   ∧ wa 104  ∀wal 1362   ∈ wcel 2164  [wsbc 2985

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 1458  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-v 2762  df-sbc 2986

This theorem is referenced by: (None)