Theorem sbimdv 2115

Description: Deduction substituting both sides of an implication, with 𝜑 and 𝑥 disjoint. See also sbimd 2284. (Contributed by Wolf Lammen, 6-May-2023.) Revise df-sb 2100. (Revised by Steven Nguyen, 6-Jul-2023.)

Hypothesis

Ref	Expression
sbimdv.1	⊢ (𝜑 → (𝜓 → 𝜒))

Assertion

Ref	Expression
sbimdv	⊢ (𝜑 → ([𝑡 / 𝑥]𝜓 → [𝑡 / 𝑥]𝜒))

Distinct variable group:   𝜑,𝑥

Allowed substitution hints:   𝜑(𝑡)   𝜓(𝑥, 𝑡)   𝜒(𝑥, 𝑡)

Proof of Theorem sbimdv

Step	Hyp	Ref	Expression
1		sbimdv.1	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
2	1	alrimiv 1960	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒))
3		spsbim 2109	. 2 ⊢ (∀𝑥(𝜓 → 𝜒) → ([𝑡 / 𝑥]𝜓 → [𝑡 / 𝑥]𝜒))
4	2, 3	syl 18	1 ⊢ (𝜑 → ([𝑡 / 𝑥]𝜓 → [𝑡 / 𝑥]𝜒))

This proof depends on syntax axioms:   → wi 4  ∀wal 1568  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100

This theorem is used by:  sbcimdv  3815  ss2abdv  4022