Metamath Proof Explorer


Theorem oppcmndclem

Description: Lemma for oppcmndc . Everything is true for two distinct elements in a singleton or an empty set (since it is impossible). Note that if this theorem and oppcendc are in -. x = y form, then both proofs should be one step shorter. (Contributed by Zhi Wang, 16-Oct-2025)

Ref Expression
Hypothesis oppcmndclem.1 φ B = A
Assertion oppcmndclem φ X B Y B X Y ψ

Proof

Step Hyp Ref Expression
1 oppcmndclem.1 φ B = A
2 df-ne X Y ¬ X = Y
3 eqeq1 x = X x = y X = y
4 eqeq2 y = Y X = y X = Y
5 mosn B = A * x x B
6 1 5 syl φ * x x B
7 moel * x x B x B y B x = y
8 6 7 sylib φ x B y B x = y
9 8 adantr φ X B Y B x B y B x = y
10 simprl φ X B Y B X B
11 simprr φ X B Y B Y B
12 3 4 9 10 11 rspc2dv φ X B Y B X = Y
13 12 pm2.24d φ X B Y B ¬ X = Y ψ
14 2 13 biimtrid φ X B Y B X Y ψ