Metamath Proof Explorer


Theorem relssresb

Description: Biconditional form of relssres , to replace it. (Contributed by BJ, 8-Nov-2026)

Ref Expression
Assertion relssresb ⊢ Rel ⁡ A ∧ dom ⁡ A ⊆ B ↔ A ↾ B = A

Proof

Step Hyp Ref Expression
1 relssdmrnb ⊢ Rel ⁡ A ↔ A ⊆ dom ⁡ A × ran ⁡ A
2 simpl ⊢ A ⊆ dom ⁡ A × ran ⁡ A ∧ dom ⁡ A ⊆ B → A ⊆ dom ⁡ A × ran ⁡ A
3 simpr ⊢ A ⊆ dom ⁡ A × ran ⁡ A ∧ dom ⁡ A ⊆ B → dom ⁡ A ⊆ B
4 ssv ⊢ ran ⁡ A ⊆ V
5 xpss12 ⊢ dom ⁡ A ⊆ B ∧ ran ⁡ A ⊆ V → dom ⁡ A × ran ⁡ A ⊆ B × V
6 3 4 5 sylancl ⊢ A ⊆ dom ⁡ A × ran ⁡ A ∧ dom ⁡ A ⊆ B → dom ⁡ A × ran ⁡ A ⊆ B × V
7 2 6 sstrd ⊢ A ⊆ dom ⁡ A × ran ⁡ A ∧ dom ⁡ A ⊆ B → A ⊆ B × V
8 1 7 sylanb ⊢ Rel ⁡ A ∧ dom ⁡ A ⊆ B → A ⊆ B × V
9 id ⊢ A ⊆ B × V → A ⊆ B × V
10 xpss ⊢ B × V ⊆ V × V
11 9 10 sstrdi ⊢ A ⊆ B × V → A ⊆ V × V
12 df-rel ⊢ Rel ⁡ A ↔ A ⊆ V × V
13 11 12 sylibr ⊢ A ⊆ B × V → Rel ⁡ A
14 dmss ⊢ A ⊆ B × V → dom ⁡ A ⊆ dom ⁡ B × V
15 dmxpss ⊢ dom ⁡ B × V ⊆ B
16 14 15 sstrdi ⊢ A ⊆ B × V → dom ⁡ A ⊆ B
17 13 16 jca ⊢ A ⊆ B × V → Rel ⁡ A ∧ dom ⁡ A ⊆ B
18 8 17 impbii ⊢ Rel ⁡ A ∧ dom ⁡ A ⊆ B ↔ A ⊆ B × V
19 bj-resssxp ⊢ A ↾ B = A ↔ A ⊆ B × V
20 18 19 bitr4i ⊢ Rel ⁡ A ∧ dom ⁡ A ⊆ B ↔ A ↾ B = A