Metamath Proof Explorer


Theorem reuf1odnf

Description: There is exactly one element in each of two isomorphic sets. Variant of reuf1od with no distinct variable condition for ch . (Contributed by AV, 19-Mar-2023)

Ref Expression
Hypotheses reuf1odnf.f φF:C1-1 ontoB
reuf1odnf.x φx=Fyψχ
reuf1odnf.z x=zψθ
reuf1odnf.n xχ
Assertion reuf1odnf φ∃!xBψ∃!yCχ

Proof

Step Hyp Ref Expression
1 reuf1odnf.f φF:C1-1 ontoB
2 reuf1odnf.x φx=Fyψχ
3 reuf1odnf.z x=zψθ
4 reuf1odnf.n xχ
5 f1of F:C1-1 ontoBF:CB
6 1 5 syl φF:CB
7 6 ffvelcdmda φyCFyB
8 f1ofveu F:C1-1 ontoBxB∃!yCFy=x
9 eqcom x=FyFy=x
10 9 reubii ∃!yCx=Fy∃!yCFy=x
11 8 10 sylibr F:C1-1 ontoBxB∃!yCx=Fy
12 1 11 sylan φxB∃!yCx=Fy
13 sbceq1a x=Fyψ[˙Fy/x]˙ψ
14 13 adantl φx=Fyψ[˙Fy/x]˙ψ
15 3 cbvsbcvw [˙Fy/x]˙ψ[˙Fy/z]˙θ
16 14 15 bitrdi φx=Fyψ[˙Fy/z]˙θ
17 7 12 16 reuxfr1d φ∃!xBψ∃!yC[˙Fy/z]˙θ
18 15 a1i φ[˙Fy/x]˙ψ[˙Fy/z]˙θ
19 18 bicomd φ[˙Fy/z]˙θ[˙Fy/x]˙ψ
20 19 reubidv φ∃!yC[˙Fy/z]˙θ∃!yC[˙Fy/x]˙ψ
21 fvexd φFyV
22 nfv xφ
23 4 a1i φxχ
24 21 2 22 23 sbciedf φ[˙Fy/x]˙ψχ
25 24 reubidv φ∃!yC[˙Fy/x]˙ψ∃!yCχ
26 17 20 25 3bitrd φ∃!xBψ∃!yCχ