Metamath Proof Explorer


Theorem isplng

Description: The property of being a plane. (Contributed by Thierry Arnoux, 17-Jun-2026)

Ref Expression
Hypotheses plngval.p ⊢ P = Base G
plngval.i ⊢ I = Itv ⁡ G
plngval.1 ⊢ L = Line 𝒢 ⁡ G
plngval.e No typesetting found for |- E = ( PlnG ` G ) with typecode |-
plngval.g ⊢ φ → G ∈ 𝒢 Tarski
isplng.h ⊢ φ → H ∈ ran ⁡ E
Assertion isplng ⊢ φ → ∃ a ∈ ran ⁡ L ∃ r ∈ P ∖ a H = a E r

Proof

Step Hyp Ref Expression
1 plngval.p ⊢ P = Base G
2 plngval.i ⊢ I = Itv ⁡ G
3 plngval.1 ⊢ L = Line 𝒢 ⁡ G
4 plngval.e Could not format E = ( PlnG ` G ) : No typesetting found for |- E = ( PlnG ` G ) with typecode |-
5 plngval.g ⊢ φ → G ∈ 𝒢 Tarski
6 isplng.h ⊢ φ → H ∈ ran ⁡ E
7 df-plng Could not format PlnG = ( g e. _V |-> ( a e. ran ( LineG ` g ) , r e. ( ( Base ` g ) \ a ) |-> { x e. ( Base ` g ) | ( x e. a \/ x ( ( hpG ` g ) ` a ) r \/ E. t e. a t e. ( x ( Itv ` g ) r ) ) } ) ) : No typesetting found for |- PlnG = ( g e. _V |-> ( a e. ran ( LineG ` g ) , r e. ( ( Base ` g ) \ a ) |-> { x e. ( Base ` g ) | ( x e. a \/ x ( ( hpG ` g ) ` a ) r \/ E. t e. a t e. ( x ( Itv ` g ) r ) ) } ) ) with typecode |-
8 fveq2 ⊢ g = G → Line 𝒢 ⁡ g = Line 𝒢 ⁡ G
9 8 3 eqtr4di ⊢ g = G → Line 𝒢 ⁡ g = L
10 9 rneqd ⊢ g = G → ran ⁡ Line 𝒢 ⁡ g = ran ⁡ L
11 fveq2 ⊢ g = G → Base g = Base G
12 11 1 eqtr4di ⊢ g = G → Base g = P
13 12 difeq1d ⊢ g = G → Base g ∖ a = P ∖ a
14 biidd ⊢ g = G → x ∈ a ↔ x ∈ a
15 fveq2 ⊢ g = G → hp 𝒢 ⁡ g = hp 𝒢 ⁡ G
16 15 fveq1d ⊢ g = G → hp 𝒢 ⁡ g ⁡ a = hp 𝒢 ⁡ G ⁡ a
17 16 breqd ⊢ g = G → x hp 𝒢 ⁡ g ⁡ a r ↔ x hp 𝒢 ⁡ G ⁡ a r
18 fveq2 ⊢ g = G → Itv ⁡ g = Itv ⁡ G
19 18 2 eqtr4di ⊢ g = G → Itv ⁡ g = I
20 19 oveqd ⊢ g = G → x Itv ⁡ g r = x I r
21 20 eleq2d ⊢ g = G → t ∈ x Itv ⁡ g r ↔ t ∈ x I r
22 21 rexbidv ⊢ g = G → ∃ t ∈ a t ∈ x Itv ⁡ g r ↔ ∃ t ∈ a t ∈ x I r
23 14 17 22 3orbi123d ⊢ g = G → x ∈ a ∨ x hp 𝒢 ⁡ g ⁡ a r ∨ ∃ t ∈ a t ∈ x Itv ⁡ g r ↔ x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
24 12 23 rabeqbidv ⊢ g = G → x ∈ Base g | x ∈ a ∨ x hp 𝒢 ⁡ g ⁡ a r ∨ ∃ t ∈ a t ∈ x Itv ⁡ g r = x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
25 10 13 24 mpoeq123dv ⊢ g = G → a ∈ ran ⁡ Line 𝒢 ⁡ g , r ∈ Base g ∖ a ⟼ x ∈ Base g | x ∈ a ∨ x hp 𝒢 ⁡ g ⁡ a r ∨ ∃ t ∈ a t ∈ x Itv ⁡ g r = a ∈ ran ⁡ L , r ∈ P ∖ a ⟼ x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
26 5 elexd ⊢ φ → G ∈ V
27 3 fvexi ⊢ L ∈ V
28 27 rnex ⊢ ran ⁡ L ∈ V
29 28 a1i ⊢ φ → ran ⁡ L ∈ V
30 1 fvexi ⊢ P ∈ V
31 30 difexi ⊢ P ∖ a ∈ V
32 31 a1i ⊢ φ ∧ a ∈ ran ⁡ L → P ∖ a ∈ V
33 29 32 mpoexd ⊢ φ → a ∈ ran ⁡ L , r ∈ P ∖ a ⟼ x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r ∈ V
34 7 25 26 33 fvmptd3 Could not format ( ph -> ( PlnG ` G ) = ( a e. ran L , r e. ( P \ a ) |-> { x e. P | ( x e. a \/ x ( ( hpG ` G ) ` a ) r \/ E. t e. a t e. ( x I r ) ) } ) ) : No typesetting found for |- ( ph -> ( PlnG ` G ) = ( a e. ran L , r e. ( P \ a ) |-> { x e. P | ( x e. a \/ x ( ( hpG ` G ) ` a ) r \/ E. t e. a t e. ( x I r ) ) } ) ) with typecode |-
35 4 34 eqtrid ⊢ φ → E = a ∈ ran ⁡ L , r ∈ P ∖ a ⟼ x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
36 35 rneqd ⊢ φ → ran ⁡ E = ran ⁡ a ∈ ran ⁡ L , r ∈ P ∖ a ⟼ x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
37 6 36 eleqtrd ⊢ φ → H ∈ ran ⁡ a ∈ ran ⁡ L , r ∈ P ∖ a ⟼ x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
38 eqid ⊢ a ∈ ran ⁡ L , r ∈ P ∖ a ⟼ x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r = a ∈ ran ⁡ L , r ∈ P ∖ a ⟼ x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
39 30 rabex ⊢ x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r ∈ V
40 38 39 elrnmpo ⊢ H ∈ ran ⁡ a ∈ ran ⁡ L , r ∈ P ∖ a ⟼ x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r ↔ ∃ a ∈ ran ⁡ L ∃ r ∈ P ∖ a H = x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
41 37 40 sylib ⊢ φ → ∃ a ∈ ran ⁡ L ∃ r ∈ P ∖ a H = x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
42 5 ad2antrr ⊢ φ ∧ a ∈ ran ⁡ L ∧ r ∈ P ∖ a → G ∈ 𝒢 Tarski
43 simplr ⊢ φ ∧ a ∈ ran ⁡ L ∧ r ∈ P ∖ a → a ∈ ran ⁡ L
44 simpr ⊢ φ ∧ a ∈ ran ⁡ L ∧ r ∈ P ∖ a → r ∈ P ∖ a
45 1 2 3 4 42 43 44 plngval ⊢ φ ∧ a ∈ ran ⁡ L ∧ r ∈ P ∖ a → a E r = x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
46 45 eqeq2d ⊢ φ ∧ a ∈ ran ⁡ L ∧ r ∈ P ∖ a → H = a E r ↔ H = x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r
47 46 biimprd ⊢ φ ∧ a ∈ ran ⁡ L ∧ r ∈ P ∖ a → H = x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r → H = a E r
48 47 anasss ⊢ φ ∧ a ∈ ran ⁡ L ∧ r ∈ P ∖ a → H = x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r → H = a E r
49 48 reximdvva ⊢ φ → ∃ a ∈ ran ⁡ L ∃ r ∈ P ∖ a H = x ∈ P | x ∈ a ∨ x hp 𝒢 ⁡ G ⁡ a r ∨ ∃ t ∈ a t ∈ x I r → ∃ a ∈ ran ⁡ L ∃ r ∈ P ∖ a H = a E r
50 41 49 mpd ⊢ φ → ∃ a ∈ ran ⁡ L ∃ r ∈ P ∖ a H = a E r