Metamath Proof Explorer


Theorem moeq3

Description: "At most one" property of equality (split into 3 cases). (The first two hypotheses could be eliminated with longer proof.) (Contributed by NM, 23-Apr-1995)

Ref Expression
Hypotheses moeq3.1 B V
moeq3.2 C V
moeq3.3 ¬ φ ψ
Assertion moeq3 * x φ x = A ¬ φ ψ x = B ψ x = C

Proof

Step Hyp Ref Expression
1 moeq3.1 B V
2 moeq3.2 C V
3 moeq3.3 ¬ φ ψ
4 eqeq2 y = A x = y x = A
5 4 anbi2d y = A φ x = y φ x = A
6 biidd y = A ¬ φ ψ x = B ¬ φ ψ x = B
7 biidd y = A ψ x = C ψ x = C
8 5 6 7 3orbi123d y = A φ x = y ¬ φ ψ x = B ψ x = C φ x = A ¬ φ ψ x = B ψ x = C
9 8 eubidv y = A ∃! x φ x = y ¬ φ ψ x = B ψ x = C ∃! x φ x = A ¬ φ ψ x = B ψ x = C
10 vex y V
11 10 1 2 3 eueq3 ∃! x φ x = y ¬ φ ψ x = B ψ x = C
12 9 11 vtoclg A V ∃! x φ x = A ¬ φ ψ x = B ψ x = C
13 eumo ∃! x φ x = A ¬ φ ψ x = B ψ x = C * x φ x = A ¬ φ ψ x = B ψ x = C
14 12 13 syl A V * x φ x = A ¬ φ ψ x = B ψ x = C
15 eqvisset x = A A V
16 pm2.21 ¬ A V A V x = y
17 15 16 syl5 ¬ A V x = A x = y
18 17 anim2d ¬ A V φ x = A φ x = y
19 18 orim1d ¬ A V φ x = A ¬ φ ψ x = B ψ x = C φ x = y ¬ φ ψ x = B ψ x = C
20 3orass φ x = A ¬ φ ψ x = B ψ x = C φ x = A ¬ φ ψ x = B ψ x = C
21 3orass φ x = y ¬ φ ψ x = B ψ x = C φ x = y ¬ φ ψ x = B ψ x = C
22 19 20 21 3imtr4g ¬ A V φ x = A ¬ φ ψ x = B ψ x = C φ x = y ¬ φ ψ x = B ψ x = C
23 22 alrimiv ¬ A V x φ x = A ¬ φ ψ x = B ψ x = C φ x = y ¬ φ ψ x = B ψ x = C
24 euimmo x φ x = A ¬ φ ψ x = B ψ x = C φ x = y ¬ φ ψ x = B ψ x = C ∃! x φ x = y ¬ φ ψ x = B ψ x = C * x φ x = A ¬ φ ψ x = B ψ x = C
25 23 11 24 mpisyl ¬ A V * x φ x = A ¬ φ ψ x = B ψ x = C
26 14 25 pm2.61i * x φ x = A ¬ φ ψ x = B ψ x = C