Metamath Proof Explorer


Theorem cdleme

Description: Lemma E in Crawley p. 113. If p,q are atoms not under hyperplane w, then there is a unique translation f such that f(p) = q. (Contributed by NM, 11-Apr-2013)

Ref Expression
Hypotheses cdleme.l ˙ = K
cdleme.a A = Atoms K
cdleme.h H = LHyp K
cdleme.t T = LTrn K W
Assertion cdleme K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W ∃! f T f P = Q

Proof

Step Hyp Ref Expression
1 cdleme.l ˙ = K
2 cdleme.a A = Atoms K
3 cdleme.h H = LHyp K
4 cdleme.t T = LTrn K W
5 1 2 3 4 cdleme50ex K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W f T f P = Q
6 simp11 K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W f T z T f P = Q z P = Q K HL W H
7 simp2l K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W f T z T f P = Q z P = Q f T
8 simp2r K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W f T z T f P = Q z P = Q z T
9 simp12 K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W f T z T f P = Q z P = Q P A ¬ P ˙ W
10 eqtr3 f P = Q z P = Q f P = z P
11 10 3ad2ant3 K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W f T z T f P = Q z P = Q f P = z P
12 1 2 3 4 cdlemd K HL W H f T z T P A ¬ P ˙ W f P = z P f = z
13 6 7 8 9 11 12 syl311anc K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W f T z T f P = Q z P = Q f = z
14 13 3exp K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W f T z T f P = Q z P = Q f = z
15 14 ralrimivv K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W f T z T f P = Q z P = Q f = z
16 fveq1 f = z f P = z P
17 16 eqeq1d f = z f P = Q z P = Q
18 17 reu4 ∃! f T f P = Q f T f P = Q f T z T f P = Q z P = Q f = z
19 5 15 18 sylanbrc K HL W H P A ¬ P ˙ W Q A ¬ Q ˙ W ∃! f T f P = Q