Metamath Proof Explorer


Theorem sbcfung

Description: Distribute proper substitution through the function predicate. (Contributed by Alexander van der Vekens, 23-Jul-2017) Shorten proof and remove dependency on ax-sep and ax-pr . (Revised by Eric Schmidt, 12-Sep-2026)

Ref Expression
Assertion sbcfung ⊢ A ∈ V → [˙A / x]˙ Fun ⁡ F ↔ Fun ⁡ ⦋ A / x⦌ F

Proof

Step Hyp Ref Expression
1 sbcan ⊢ [˙A / x]˙ Rel ⁡ F ∧ F ∘ F -1 ⊆ I ↔ [˙A / x]˙ Rel ⁡ F ∧ [˙A / x]˙ F ∘ F -1 ⊆ I
2 sbcrel ⊢ A ∈ V → [˙A / x]˙ Rel ⁡ F ↔ Rel ⁡ ⦋ A / x⦌ F
3 sbcssg ⊢ A ∈ V → [˙A / x]˙ F ∘ F -1 ⊆ I ↔ ⦋ A / x⦌ F ∘ F -1 ⊆ ⦋ A / x⦌ I
4 csbcog ⊢ A ∈ V → ⦋ A / x⦌ F ∘ F -1 = ⦋ A / x⦌ F ∘ ⦋ A / x⦌ F -1
5 csbcnv ⊢ ⦋ A / x⦌ F -1 = ⦋ A / x⦌ F -1
6 5 coeq2i ⊢ ⦋ A / x⦌ F ∘ ⦋ A / x⦌ F -1 = ⦋ A / x⦌ F ∘ ⦋ A / x⦌ F -1
7 4 6 eqtr4di ⊢ A ∈ V → ⦋ A / x⦌ F ∘ F -1 = ⦋ A / x⦌ F ∘ ⦋ A / x⦌ F -1
8 csbconstg ⊢ A ∈ V → ⦋ A / x⦌ I = I
9 7 8 sseq12d ⊢ A ∈ V → ⦋ A / x⦌ F ∘ F -1 ⊆ ⦋ A / x⦌ I ↔ ⦋ A / x⦌ F ∘ ⦋ A / x⦌ F -1 ⊆ I
10 3 9 bitrd ⊢ A ∈ V → [˙A / x]˙ F ∘ F -1 ⊆ I ↔ ⦋ A / x⦌ F ∘ ⦋ A / x⦌ F -1 ⊆ I
11 2 10 anbi12d ⊢ A ∈ V → [˙A / x]˙ Rel ⁡ F ∧ [˙A / x]˙ F ∘ F -1 ⊆ I ↔ Rel ⁡ ⦋ A / x⦌ F ∧ ⦋ A / x⦌ F ∘ ⦋ A / x⦌ F -1 ⊆ I
12 1 11 bitrid ⊢ A ∈ V → [˙A / x]˙ Rel ⁡ F ∧ F ∘ F -1 ⊆ I ↔ Rel ⁡ ⦋ A / x⦌ F ∧ ⦋ A / x⦌ F ∘ ⦋ A / x⦌ F -1 ⊆ I
13 df-fun ⊢ Fun ⁡ F ↔ Rel ⁡ F ∧ F ∘ F -1 ⊆ I
14 13 sbcbii ⊢ [˙A / x]˙ Fun ⁡ F ↔ [˙A / x]˙ Rel ⁡ F ∧ F ∘ F -1 ⊆ I
15 df-fun ⊢ Fun ⁡ ⦋ A / x⦌ F ↔ Rel ⁡ ⦋ A / x⦌ F ∧ ⦋ A / x⦌ F ∘ ⦋ A / x⦌ F -1 ⊆ I
16 12 14 15 3bitr4g ⊢ A ∈ V → [˙A / x]˙ Fun ⁡ F ↔ Fun ⁡ ⦋ A / x⦌ F