Metamath Proof Explorer


Theorem dvelimhw

Description: Proof of dvelimh without using ax-13 but with additional distinct variable conditions. (Contributed by NM, 1-Oct-2002) (Revised by Andrew Salmon, 21-Jul-2011) (Revised by NM, 1-Aug-2017) (Proof shortened by Wolf Lammen, 23-Dec-2018)

Ref Expression
Hypotheses dvelimhw.1 φ x φ
dvelimhw.2 ψ z ψ
dvelimhw.3 z = y φ ψ
dvelimhw.4 ¬ x x = y y = z x y = z
Assertion dvelimhw ¬ x x = y ψ x ψ

Proof

Step Hyp Ref Expression
1 dvelimhw.1 φ x φ
2 dvelimhw.2 ψ z ψ
3 dvelimhw.3 z = y φ ψ
4 dvelimhw.4 ¬ x x = y y = z x y = z
5 nfv z ¬ x x = y
6 equcom z = y y = z
7 nfna1 x ¬ x x = y
8 7 4 nf5d ¬ x x = y x y = z
9 6 8 nfxfrd ¬ x x = y x z = y
10 1 nf5i x φ
11 10 a1i ¬ x x = y x φ
12 9 11 nfimd ¬ x x = y x z = y φ
13 5 12 nfald ¬ x x = y x z z = y φ
14 2 3 equsalhw z z = y φ ψ
15 14 nfbii x z z = y φ x ψ
16 13 15 sylib ¬ x x = y x ψ
17 16 nf5rd ¬ x x = y ψ x ψ