Metamath Proof Explorer


Theorem setrecsss

Description: The setrecs operator respects the subset relation between two functions F and G . (Contributed by Emmett Weisz, 13-Mar-2022)

Ref Expression
Hypotheses setrecsss.1 ⊢ φ → Fun ⁡ G
setrecsss.2 ⊢ φ → F ⊆ G
Assertion setrecsss ⊢ φ → setrecs ⁡ F ⊆ setrecs ⁡ G

Proof

Step Hyp Ref Expression
1 setrecsss.1 ⊢ φ → Fun ⁡ G
2 setrecsss.2 ⊢ φ → F ⊆ G
3 eqid ⊢ setrecs ⁡ F = setrecs ⁡ F
4 imass1 ⊢ F ⊆ G → F x ⊆ G x
5 2 4 syl ⊢ φ → F x ⊆ G x
6 5 unissd ⊢ φ → ⋃ F x ⊆ ⋃ G x
7 funss ⊢ F ⊆ G → Fun ⁡ G → Fun ⁡ F
8 2 1 7 sylc ⊢ φ → Fun ⁡ F
9 funfv ⊢ Fun ⁡ F → F ⁡ x = ⋃ F x
10 8 9 syl ⊢ φ → F ⁡ x = ⋃ F x
11 funfv ⊢ Fun ⁡ G → G ⁡ x = ⋃ G x
12 1 11 syl ⊢ φ → G ⁡ x = ⋃ G x
13 6 10 12 3sstr4d ⊢ φ → F ⁡ x ⊆ G ⁡ x
14 13 adantr ⊢ φ ∧ x ⊆ setrecs ⁡ G → F ⁡ x ⊆ G ⁡ x
15 eqid ⊢ setrecs ⁡ G = setrecs ⁡ G
16 vex ⊢ x ∈ V
17 16 a1i ⊢ φ ∧ x ⊆ setrecs ⁡ G → x ∈ V
18 simpr ⊢ φ ∧ x ⊆ setrecs ⁡ G → x ⊆ setrecs ⁡ G
19 15 17 18 setrec1 ⊢ φ ∧ x ⊆ setrecs ⁡ G → G ⁡ x ⊆ setrecs ⁡ G
20 14 19 sstrd ⊢ φ ∧ x ⊆ setrecs ⁡ G → F ⁡ x ⊆ setrecs ⁡ G
21 20 ex ⊢ φ → x ⊆ setrecs ⁡ G → F ⁡ x ⊆ setrecs ⁡ G
22 21 alrimiv ⊢ φ → ∀ x x ⊆ setrecs ⁡ G → F ⁡ x ⊆ setrecs ⁡ G
23 3 22 setrec2v ⊢ φ → setrecs ⁡ F ⊆ setrecs ⁡ G