Metamath Proof Explorer


Theorem sbcom2

Description: Commutativity law for substitution. Used in proof of Theorem 9.7 of Megill p. 449 (p. 16 of the preprint). (Contributed by NM, 27-May-1997) (Proof shortened by Wolf Lammen, 23-Dec-2022)

Ref Expression
Assertion sbcom2 w z y x φ y x w z φ

Proof

Step Hyp Ref Expression
1 2sb6 v z u x φ z x z = v x = u φ
2 alcom z x z = v x = u φ x z z = v x = u φ
3 ancomst z = v x = u φ x = u z = v φ
4 3 2albii x z z = v x = u φ x z x = u z = v φ
5 1 2 4 3bitri v z u x φ x z x = u z = v φ
6 2sb6 u x v z φ x z x = u z = v φ
7 5 6 bitr4i v z u x φ u x v z φ
8 sbequ u = y u x φ y x φ
9 8 sbbidv u = y v z u x φ v z y x φ
10 7 9 bitr3id u = y u x v z φ v z y x φ
11 sbequ v = w v z y x φ w z y x φ
12 10 11 sylan9bb u = y v = w u x v z φ w z y x φ
13 sbequ v = w v z φ w z φ
14 13 sbbidv v = w u x v z φ u x w z φ
15 sbequ u = y u x w z φ y x w z φ
16 14 15 sylan9bbr u = y v = w u x v z φ y x w z φ
17 12 16 bitr3d u = y v = w w z y x φ y x w z φ
18 17 ex u = y v = w w z y x φ y x w z φ
19 ax6ev u u = y
20 18 19 exlimiiv v = w w z y x φ y x w z φ
21 ax6ev v v = w
22 20 21 exlimiiv w z y x φ y x w z φ