Metamath Proof Explorer


Theorem pmtrdifellem1

Description: Lemma 1 for pmtrdifel . (Contributed by AV, 15-Jan-2019)

Ref Expression
Hypotheses pmtrdifel.t ⊢ T = ran ⁡ pmTrsp ⁡ N ∖ K
pmtrdifel.r ⊢ R = ran ⁡ pmTrsp ⁡ N
pmtrdifel.0 ⊢ S = pmTrsp ⁡ N ⁡ dom ⁡ Q ∖ I
Assertion pmtrdifellem1 ⊢ Q ∈ T → S ∈ R

Proof

Step Hyp Ref Expression
1 pmtrdifel.t ⊢ T = ran ⁡ pmTrsp ⁡ N ∖ K
2 pmtrdifel.r ⊢ R = ran ⁡ pmTrsp ⁡ N
3 pmtrdifel.0 ⊢ S = pmTrsp ⁡ N ⁡ dom ⁡ Q ∖ I
4 eqid ⊢ pmTrsp ⁡ N ∖ K = pmTrsp ⁡ N ∖ K
5 4 1 pmtrfb ⊢ Q ∈ T ↔ N ∖ K ∈ V ∧ Q : N ∖ K ⟶ 1-1 onto N ∖ K ∧ dom ⁡ Q ∖ I ≈ 2 𝑜
6 difsnexi ⊢ N ∖ K ∈ V → N ∈ V
7 f1of ⊢ Q : N ∖ K ⟶ 1-1 onto N ∖ K → Q : N ∖ K ⟶ N ∖ K
8 fdm ⊢ Q : N ∖ K ⟶ N ∖ K → dom ⁡ Q = N ∖ K
9 difssd ⊢ dom ⁡ Q = N ∖ K → Q ∖ I ⊆ Q
10 dmss ⊢ Q ∖ I ⊆ Q → dom ⁡ Q ∖ I ⊆ dom ⁡ Q
11 9 10 syl ⊢ dom ⁡ Q = N ∖ K → dom ⁡ Q ∖ I ⊆ dom ⁡ Q
12 difssd ⊢ dom ⁡ Q = N ∖ K → N ∖ K ⊆ N
13 sseq1 ⊢ dom ⁡ Q = N ∖ K → dom ⁡ Q ⊆ N ↔ N ∖ K ⊆ N
14 12 13 mpbird ⊢ dom ⁡ Q = N ∖ K → dom ⁡ Q ⊆ N
15 11 14 sstrd ⊢ dom ⁡ Q = N ∖ K → dom ⁡ Q ∖ I ⊆ N
16 7 8 15 3syl ⊢ Q : N ∖ K ⟶ 1-1 onto N ∖ K → dom ⁡ Q ∖ I ⊆ N
17 id ⊢ dom ⁡ Q ∖ I ≈ 2 𝑜 → dom ⁡ Q ∖ I ≈ 2 𝑜
18 eqid ⊢ pmTrsp ⁡ N = pmTrsp ⁡ N
19 18 2 pmtrrn ⊢ N ∈ V ∧ dom ⁡ Q ∖ I ⊆ N ∧ dom ⁡ Q ∖ I ≈ 2 𝑜 → pmTrsp ⁡ N ⁡ dom ⁡ Q ∖ I ∈ R
20 3 19 eqeltrid ⊢ N ∈ V ∧ dom ⁡ Q ∖ I ⊆ N ∧ dom ⁡ Q ∖ I ≈ 2 𝑜 → S ∈ R
21 6 16 17 20 syl3an ⊢ N ∖ K ∈ V ∧ Q : N ∖ K ⟶ 1-1 onto N ∖ K ∧ dom ⁡ Q ∖ I ≈ 2 𝑜 → S ∈ R
22 5 21 sylbi ⊢ Q ∈ T → S ∈ R