Metamath Proof Explorer


Theorem sbcco

Description: A composition law for class substitution. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker sbccow when possible. (Contributed by NM, 26-Sep-2003) (Revised by Mario Carneiro, 13-Oct-2016) (New usage is discouraged.)

Ref Expression
Assertion sbcco [˙A/y]˙[˙y/x]˙φ[˙A/x]˙φ

Proof

Step Hyp Ref Expression
1 sbcex [˙A/y]˙[˙y/x]˙φAV
2 sbcex [˙A/x]˙φAV
3 dfsbcq z=A[˙z/y]˙[˙y/x]˙φ[˙A/y]˙[˙y/x]˙φ
4 dfsbcq z=A[˙z/x]˙φ[˙A/x]˙φ
5 sbsbc yxφ[˙y/x]˙φ
6 5 sbbii zyyxφzy[˙y/x]˙φ
7 nfv yφ
8 7 sbco2 zyyxφzxφ
9 sbsbc zy[˙y/x]˙φ[˙z/y]˙[˙y/x]˙φ
10 6 8 9 3bitr3ri [˙z/y]˙[˙y/x]˙φzxφ
11 sbsbc zxφ[˙z/x]˙φ
12 10 11 bitri [˙z/y]˙[˙y/x]˙φ[˙z/x]˙φ
13 3 4 12 vtoclbg AV[˙A/y]˙[˙y/x]˙φ[˙A/x]˙φ
14 1 2 13 pm5.21nii [˙A/y]˙[˙y/x]˙φ[˙A/x]˙φ