Metamath Proof Explorer


Theorem fpwwecbv

Description: Lemma for fpwwe . (Contributed by Mario Carneiro, 15-May-2015)

Ref Expression
Hypothesis fpwwe.1 ⊢ W = x r | x ⊆ A ∧ r ⊆ x × x ∧ r We x ∧ ∀ y ∈ x F ⁡ r -1 y = y
Assertion fpwwecbv ⊢ W = a s | a ⊆ A ∧ s ⊆ a × a ∧ s We a ∧ ∀ z ∈ a F ⁡ s -1 z = z

Proof

Step Hyp Ref Expression
1 fpwwe.1 ⊢ W = x r | x ⊆ A ∧ r ⊆ x × x ∧ r We x ∧ ∀ y ∈ x F ⁡ r -1 y = y
2 simpl ⊢ x = a ∧ r = s → x = a
3 2 sseq1d ⊢ x = a ∧ r = s → x ⊆ A ↔ a ⊆ A
4 simpr ⊢ x = a ∧ r = s → r = s
5 2 sqxpeqd ⊢ x = a ∧ r = s → x × x = a × a
6 4 5 sseq12d ⊢ x = a ∧ r = s → r ⊆ x × x ↔ s ⊆ a × a
7 3 6 anbi12d ⊢ x = a ∧ r = s → x ⊆ A ∧ r ⊆ x × x ↔ a ⊆ A ∧ s ⊆ a × a
8 4 2 weeq12d ⊢ x = a ∧ r = s → r We x ↔ s We a
9 sneq ⊢ y = z → y = z
10 9 imaeq2d ⊢ y = z → r -1 y = r -1 z
11 10 fveq2d ⊢ y = z → F ⁡ r -1 y = F ⁡ r -1 z
12 id ⊢ y = z → y = z
13 11 12 eqeq12d ⊢ y = z → F ⁡ r -1 y = y ↔ F ⁡ r -1 z = z
14 13 cbvralvw ⊢ ∀ y ∈ x F ⁡ r -1 y = y ↔ ∀ z ∈ x F ⁡ r -1 z = z
15 4 cnveqd ⊢ x = a ∧ r = s → r -1 = s -1
16 15 imaeq1d ⊢ x = a ∧ r = s → r -1 z = s -1 z
17 16 fveqeq2d ⊢ x = a ∧ r = s → F ⁡ r -1 z = z ↔ F ⁡ s -1 z = z
18 2 17 raleqbidv ⊢ x = a ∧ r = s → ∀ z ∈ x F ⁡ r -1 z = z ↔ ∀ z ∈ a F ⁡ s -1 z = z
19 14 18 bitrid ⊢ x = a ∧ r = s → ∀ y ∈ x F ⁡ r -1 y = y ↔ ∀ z ∈ a F ⁡ s -1 z = z
20 8 19 anbi12d ⊢ x = a ∧ r = s → r We x ∧ ∀ y ∈ x F ⁡ r -1 y = y ↔ s We a ∧ ∀ z ∈ a F ⁡ s -1 z = z
21 7 20 anbi12d ⊢ x = a ∧ r = s → x ⊆ A ∧ r ⊆ x × x ∧ r We x ∧ ∀ y ∈ x F ⁡ r -1 y = y ↔ a ⊆ A ∧ s ⊆ a × a ∧ s We a ∧ ∀ z ∈ a F ⁡ s -1 z = z
22 21 cbvopabv ⊢ x r | x ⊆ A ∧ r ⊆ x × x ∧ r We x ∧ ∀ y ∈ x F ⁡ r -1 y = y = a s | a ⊆ A ∧ s ⊆ a × a ∧ s We a ∧ ∀ z ∈ a F ⁡ s -1 z = z
23 1 22 eqtri ⊢ W = a s | a ⊆ A ∧ s ⊆ a × a ∧ s We a ∧ ∀ z ∈ a F ⁡ s -1 z = z