Metamath Proof Explorer


Theorem relpfrlem

Description: Lemma for relpfr . Proved without using the Axiom of Replacement. This is isofrlem with weaker hypotheses. (Contributed by Eric Schmidt, 11-Oct-2025)

Ref Expression
Hypotheses relpfrlem.1 No typesetting found for |- ( ph -> H RelPres R , S ( A , B ) ) with typecode |-
relpfrlem.2 ⊢ φ → H x ∈ V
Assertion relpfrlem ⊢ φ → S Fr B → R Fr A

Proof

Step Hyp Ref Expression
1 relpfrlem.1 Could not format ( ph -> H RelPres R , S ( A , B ) ) : No typesetting found for |- ( ph -> H RelPres R , S ( A , B ) ) with typecode |-
2 relpfrlem.2 ⊢ φ → H x ∈ V
3 relpf Could not format ( H RelPres R , S ( A , B ) -> H : A --> B ) : No typesetting found for |- ( H RelPres R , S ( A , B ) -> H : A --> B ) with typecode |-
4 1 3 syl ⊢ φ → H : A ⟶ B
5 ffn ⊢ H : A ⟶ B → H Fn A
6 n0 ⊢ x ≠ ∅ ↔ ∃ y y ∈ x
7 fnfvima ⊢ H Fn A ∧ x ⊆ A ∧ y ∈ x → H ⁡ y ∈ H x
8 7 ne0d ⊢ H Fn A ∧ x ⊆ A ∧ y ∈ x → H x ≠ ∅
9 8 3expia ⊢ H Fn A ∧ x ⊆ A → y ∈ x → H x ≠ ∅
10 9 exlimdv ⊢ H Fn A ∧ x ⊆ A → ∃ y y ∈ x → H x ≠ ∅
11 6 10 biimtrid ⊢ H Fn A ∧ x ⊆ A → x ≠ ∅ → H x ≠ ∅
12 11 expimpd ⊢ H Fn A → x ⊆ A ∧ x ≠ ∅ → H x ≠ ∅
13 5 12 syl ⊢ H : A ⟶ B → x ⊆ A ∧ x ≠ ∅ → H x ≠ ∅
14 fimass ⊢ H : A ⟶ B → H x ⊆ B
15 13 14 jctild ⊢ H : A ⟶ B → x ⊆ A ∧ x ≠ ∅ → H x ⊆ B ∧ H x ≠ ∅
16 4 15 syl ⊢ φ → x ⊆ A ∧ x ≠ ∅ → H x ⊆ B ∧ H x ≠ ∅
17 dffr3 ⊢ S Fr B ↔ ∀ z z ⊆ B ∧ z ≠ ∅ → ∃ w ∈ z z ∩ S -1 w = ∅
18 sseq1 ⊢ z = H x → z ⊆ B ↔ H x ⊆ B
19 neeq1 ⊢ z = H x → z ≠ ∅ ↔ H x ≠ ∅
20 18 19 anbi12d ⊢ z = H x → z ⊆ B ∧ z ≠ ∅ ↔ H x ⊆ B ∧ H x ≠ ∅
21 ineq1 ⊢ z = H x → z ∩ S -1 w = H x ∩ S -1 w
22 21 eqeq1d ⊢ z = H x → z ∩ S -1 w = ∅ ↔ H x ∩ S -1 w = ∅
23 22 rexeqbi1dv ⊢ z = H x → ∃ w ∈ z z ∩ S -1 w = ∅ ↔ ∃ w ∈ H x H x ∩ S -1 w = ∅
24 20 23 imbi12d ⊢ z = H x → z ⊆ B ∧ z ≠ ∅ → ∃ w ∈ z z ∩ S -1 w = ∅ ↔ H x ⊆ B ∧ H x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅
25 24 spcgv ⊢ H x ∈ V → ∀ z z ⊆ B ∧ z ≠ ∅ → ∃ w ∈ z z ∩ S -1 w = ∅ → H x ⊆ B ∧ H x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅
26 2 25 syl ⊢ φ → ∀ z z ⊆ B ∧ z ≠ ∅ → ∃ w ∈ z z ∩ S -1 w = ∅ → H x ⊆ B ∧ H x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅
27 17 26 biimtrid ⊢ φ → S Fr B → H x ⊆ B ∧ H x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅
28 16 27 syl5d ⊢ φ → S Fr B → x ⊆ A ∧ x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅
29 4 adantr ⊢ φ ∧ x ⊆ A → H : A ⟶ B
30 29 ffund ⊢ φ ∧ x ⊆ A → Fun ⁡ H
31 simpl ⊢ w ∈ H x ∧ H x ∩ S -1 w = ∅ → w ∈ H x
32 fvelima ⊢ Fun ⁡ H ∧ w ∈ H x → ∃ y ∈ x H ⁡ y = w
33 30 31 32 syl2an ⊢ φ ∧ x ⊆ A ∧ w ∈ H x ∧ H x ∩ S -1 w = ∅ → ∃ y ∈ x H ⁡ y = w
34 sneq ⊢ w = H ⁡ y → w = H ⁡ y
35 34 eqcoms ⊢ H ⁡ y = w → w = H ⁡ y
36 35 imaeq2d ⊢ H ⁡ y = w → S -1 w = S -1 H ⁡ y
37 36 ineq2d ⊢ H ⁡ y = w → H x ∩ S -1 w = H x ∩ S -1 H ⁡ y
38 37 eqeq1d ⊢ H ⁡ y = w → H x ∩ S -1 w = ∅ ↔ H x ∩ S -1 H ⁡ y = ∅
39 38 biimpd ⊢ H ⁡ y = w → H x ∩ S -1 w = ∅ → H x ∩ S -1 H ⁡ y = ∅
40 ssel ⊢ x ⊆ A → y ∈ x → y ∈ A
41 40 imdistani ⊢ x ⊆ A ∧ y ∈ x → x ⊆ A ∧ y ∈ A
42 relpmin Could not format ( ( H RelPres R , S ( A , B ) /\ ( x C_ A /\ y e. A ) ) -> ( ( ( H " x ) i^i ( `' S " { ( H ` y ) } ) ) = (/) -> ( x i^i ( `' R " { y } ) ) = (/) ) ) : No typesetting found for |- ( ( H RelPres R , S ( A , B ) /\ ( x C_ A /\ y e. A ) ) -> ( ( ( H " x ) i^i ( `' S " { ( H ` y ) } ) ) = (/) -> ( x i^i ( `' R " { y } ) ) = (/) ) ) with typecode |-
43 1 41 42 syl2an ⊢ φ ∧ x ⊆ A ∧ y ∈ x → H x ∩ S -1 H ⁡ y = ∅ → x ∩ R -1 y = ∅
44 39 43 sylan9r ⊢ φ ∧ x ⊆ A ∧ y ∈ x ∧ H ⁡ y = w → H x ∩ S -1 w = ∅ → x ∩ R -1 y = ∅
45 44 adantld ⊢ φ ∧ x ⊆ A ∧ y ∈ x ∧ H ⁡ y = w → w ∈ H x ∧ H x ∩ S -1 w = ∅ → x ∩ R -1 y = ∅
46 45 exp42 ⊢ φ → x ⊆ A → y ∈ x → H ⁡ y = w → w ∈ H x ∧ H x ∩ S -1 w = ∅ → x ∩ R -1 y = ∅
47 46 imp ⊢ φ ∧ x ⊆ A → y ∈ x → H ⁡ y = w → w ∈ H x ∧ H x ∩ S -1 w = ∅ → x ∩ R -1 y = ∅
48 47 com3l ⊢ y ∈ x → H ⁡ y = w → φ ∧ x ⊆ A → w ∈ H x ∧ H x ∩ S -1 w = ∅ → x ∩ R -1 y = ∅
49 48 com4t ⊢ φ ∧ x ⊆ A → w ∈ H x ∧ H x ∩ S -1 w = ∅ → y ∈ x → H ⁡ y = w → x ∩ R -1 y = ∅
50 49 imp ⊢ φ ∧ x ⊆ A ∧ w ∈ H x ∧ H x ∩ S -1 w = ∅ → y ∈ x → H ⁡ y = w → x ∩ R -1 y = ∅
51 50 reximdvai ⊢ φ ∧ x ⊆ A ∧ w ∈ H x ∧ H x ∩ S -1 w = ∅ → ∃ y ∈ x H ⁡ y = w → ∃ y ∈ x x ∩ R -1 y = ∅
52 33 51 mpd ⊢ φ ∧ x ⊆ A ∧ w ∈ H x ∧ H x ∩ S -1 w = ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
53 52 rexlimdvaa ⊢ φ ∧ x ⊆ A → ∃ w ∈ H x H x ∩ S -1 w = ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
54 53 ex ⊢ φ → x ⊆ A → ∃ w ∈ H x H x ∩ S -1 w = ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
55 54 adantrd ⊢ φ → x ⊆ A ∧ x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
56 55 a2d ⊢ φ → x ⊆ A ∧ x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅ → x ⊆ A ∧ x ≠ ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
57 28 56 syld ⊢ φ → S Fr B → x ⊆ A ∧ x ≠ ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
58 57 alrimdv ⊢ φ → S Fr B → ∀ x x ⊆ A ∧ x ≠ ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
59 dffr3 ⊢ R Fr A ↔ ∀ x x ⊆ A ∧ x ≠ ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
60 58 59 imbitrrdi ⊢ φ → S Fr B → R Fr A