Metamath Proof Explorer


Theorem preimane

Description: Different elements have different preimages. (Contributed by Thierry Arnoux, 7-May-2023)

Ref Expression
Hypotheses preimane.f ⊢ φ → Fun ⁡ F
preimane.x ⊢ φ → X ≠ Y
preimane.y ⊢ φ → X ∈ ran ⁡ F
preimane.1 ⊢ φ → Y ∈ ran ⁡ F
Assertion preimane ⊢ φ → F -1 X ≠ F -1 Y

Proof

Step Hyp Ref Expression
1 preimane.f ⊢ φ → Fun ⁡ F
2 preimane.x ⊢ φ → X ≠ Y
3 preimane.y ⊢ φ → X ∈ ran ⁡ F
4 preimane.1 ⊢ φ → Y ∈ ran ⁡ F
5 sneqrg ⊢ X ∈ ran ⁡ F → X = Y → X = Y
6 3 5 syl ⊢ φ → X = Y → X = Y
7 6 necon3d ⊢ φ → X ≠ Y → X ≠ Y
8 2 7 mpd ⊢ φ → X ≠ Y
9 funimacnv ⊢ Fun ⁡ F → F F -1 X = X ∩ ran ⁡ F
10 1 9 syl ⊢ φ → F F -1 X = X ∩ ran ⁡ F
11 3 snssd ⊢ φ → X ⊆ ran ⁡ F
12 dfss2 ⊢ X ⊆ ran ⁡ F ↔ X ∩ ran ⁡ F = X
13 11 12 sylib ⊢ φ → X ∩ ran ⁡ F = X
14 10 13 eqtrd ⊢ φ → F F -1 X = X
15 funimacnv ⊢ Fun ⁡ F → F F -1 Y = Y ∩ ran ⁡ F
16 1 15 syl ⊢ φ → F F -1 Y = Y ∩ ran ⁡ F
17 4 snssd ⊢ φ → Y ⊆ ran ⁡ F
18 dfss2 ⊢ Y ⊆ ran ⁡ F ↔ Y ∩ ran ⁡ F = Y
19 17 18 sylib ⊢ φ → Y ∩ ran ⁡ F = Y
20 16 19 eqtrd ⊢ φ → F F -1 Y = Y
21 8 14 20 3netr4d ⊢ φ → F F -1 X ≠ F F -1 Y
22 imaeq2 ⊢ F -1 X = F -1 Y → F F -1 X = F F -1 Y
23 22 necon3i ⊢ F F -1 X ≠ F F -1 Y → F -1 X ≠ F -1 Y
24 21 23 syl ⊢ φ → F -1 X ≠ F -1 Y