cdlemkid5

	Ref	Expression
Hypotheses	cdlemk5.b	$⊢ B = {Base}_{K}$
	cdlemk5.l	$⊢ \underset{˙}{\leq} = \leq_{K}$
	cdlemk5.j	$⊢ \underset{˙}{\lor} = join (K)$
	cdlemk5.m	$⊢ \underset{˙}{\land} = meet (K)$
	cdlemk5.a	$⊢ A = Atoms (K)$
	cdlemk5.h	$⊢ H = LHyp (K)$
	cdlemk5.t	$⊢ T = LTrn (K) (W)$
	cdlemk5.r	$⊢ R = trL (K) (W)$
	cdlemk5.z	$⊢ Z = (P \underset{˙}{\lor} R (b)) \underset{˙}{\land} (N (P) \underset{˙}{\lor} R (b \circ F^{-1}))$
	cdlemk5.y	$⊢ Y = (P \underset{˙}{\lor} R (g)) \underset{˙}{\land} (Z \underset{˙}{\lor} R (g \circ b^{-1}))$
	cdlemk5.x	$⊢ X = (ι z \in T \| \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (g)) \to z (P) = Y))$
Assertion	cdlemkid5	$⊢ ((K \in HL \land W \in H) \land (F \in T \land N \in T \land R (F) = R (N)) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land G = {I ↾}_{B})) \to ⦋ G / g ⦌ X \in T$

Proof

Step	Hyp	Ref	Expression
1		cdlemk5.b	$⊢ B = {Base}_{K}$
2		cdlemk5.l	$⊢ \underset{˙}{\leq} = \leq_{K}$
3		cdlemk5.j	$⊢ \underset{˙}{\lor} = join (K)$
4		cdlemk5.m	$⊢ \underset{˙}{\land} = meet (K)$
5		cdlemk5.a	$⊢ A = Atoms (K)$
6		cdlemk5.h	$⊢ H = LHyp (K)$
7		cdlemk5.t	$⊢ T = LTrn (K) (W)$
8		cdlemk5.r	$⊢ R = trL (K) (W)$
9		cdlemk5.z	$⊢ Z = (P \underset{˙}{\lor} R (b)) \underset{˙}{\land} (N (P) \underset{˙}{\lor} R (b \circ F^{-1}))$
10		cdlemk5.y	$⊢ Y = (P \underset{˙}{\lor} R (g)) \underset{˙}{\land} (Z \underset{˙}{\lor} R (g \circ b^{-1}))$
11		cdlemk5.x	$⊢ X = (ι z \in T \| \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (g)) \to z (P) = Y))$
12	1 2 3 4 5 6 7 8 9 10 11	cdlemkid4	$⊢ ((K \in HL \land W \in H) \land (F \in T \land N \in T \land R (F) = R (N)) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land G = {I ↾}_{B})) \to ⦋ G / g ⦌ X = (ι z \in T \| \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B}))$
13	1 6 7	idltrn	$⊢ (K \in HL \land W \in H) \to {I ↾}_{B} \in T$
14	13	3ad2ant1	$⊢ ((K \in HL \land W \in H) \land (F \in T \land N \in T \land R (F) = R (N)) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land G = {I ↾}_{B})) \to {I ↾}_{B} \in T$
15		eqidd	$⊢ (b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to {I ↾}_{B} = {I ↾}_{B}$
16	15	rgenw	$⊢ \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to {I ↾}_{B} = {I ↾}_{B})$
17		eqeq1	$⊢ z = {I ↾}_{B} \to (z = {I ↾}_{B} \leftrightarrow {I ↾}_{B} = {I ↾}_{B})$
18	17	imbi2d	$⊢ z = {I ↾}_{B} \to (((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B}) \leftrightarrow ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to {I ↾}_{B} = {I ↾}_{B}))$
19	18	ralbidv	$⊢ z = {I ↾}_{B} \to (\forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B}) \leftrightarrow \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to {I ↾}_{B} = {I ↾}_{B}))$
20	19	rspcev	$⊢ ({I ↾}_{B} \in T \land \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to {I ↾}_{B} = {I ↾}_{B})) \to \exists z \in T \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B})$
21	14 16 20	sylancl	$⊢ ((K \in HL \land W \in H) \land (F \in T \land N \in T \land R (F) = R (N)) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land G = {I ↾}_{B})) \to \exists z \in T \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B})$
22	1 6 7 8	cdlemftr2	$⊢ (K \in HL \land W \in H) \to \exists b \in T (b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G))$
23	22	3ad2ant1	$⊢ ((K \in HL \land W \in H) \land (F \in T \land N \in T \land R (F) = R (N)) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land G = {I ↾}_{B})) \to \exists b \in T (b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G))$
24		reusv1	$⊢ \exists b \in T (b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to (\exists! z \in T \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B}) \leftrightarrow \exists z \in T \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B}))$
25	23 24	syl	$⊢ ((K \in HL \land W \in H) \land (F \in T \land N \in T \land R (F) = R (N)) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land G = {I ↾}_{B})) \to (\exists! z \in T \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B}) \leftrightarrow \exists z \in T \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B}))$
26	21 25	mpbird	$⊢ ((K \in HL \land W \in H) \land (F \in T \land N \in T \land R (F) = R (N)) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land G = {I ↾}_{B})) \to \exists! z \in T \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B})$
27		riotacl	$⊢ \exists! z \in T \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B}) \to (ι z \in T \| \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B})) \in T$
28	26 27	syl	$⊢ ((K \in HL \land W \in H) \land (F \in T \land N \in T \land R (F) = R (N)) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land G = {I ↾}_{B})) \to (ι z \in T \| \forall b \in T ((b \neq {I ↾}_{B} \land R (b) \neq R (F) \land R (b) \neq R (G)) \to z = {I ↾}_{B})) \in T$
29	12 28	eqeltrd	$⊢ ((K \in HL \land W \in H) \land (F \in T \land N \in T \land R (F) = R (N)) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land G = {I ↾}_{B})) \to ⦋ G / g ⦌ X \in T$

Metamath Proof Explorer

Theorem cdlemkid5

Proof