Metamath Proof Explorer

Theorem cdlemg2dN

Description: This theorem can be used to shorten G = hypothesis. TODO: Fix comment. (Contributed by NM, 21-Apr-2013) (New usage is discouraged.)

		Ref	Expression
	Hypotheses	cdlemg2.b	$⊢ B = {Base}_{K}$
		cdlemg2.l	$⊢ \underset{˙}{\leq} = \leq_{K}$
		cdlemg2.j	$⊢ \underset{˙}{\lor} = join (K)$
		cdlemg2.m	$⊢ \underset{˙}{\land} = meet (K)$
		cdlemg2.a	$⊢ A = Atoms (K)$
		cdlemg2.h	$⊢ H = LHyp (K)$
		cdlemg2.t	$⊢ T = LTrn (K) (W)$
		cdlemg2.u	$⊢ U = (P \underset{˙}{\lor} Q) \underset{˙}{\land} W$
		cdlemg2.d	$⊢ D = (t \underset{˙}{\lor} U) \underset{˙}{\land} (Q \underset{˙}{\lor} ((P \underset{˙}{\lor} t) \underset{˙}{\land} W))$
		cdlemg2.e	$⊢ E = (P \underset{˙}{\lor} Q) \underset{˙}{\land} (D \underset{˙}{\lor} ((s \underset{˙}{\lor} t) \underset{˙}{\land} W))$
		cdlemg2.g	$⊢ G = (x \in B ⟼ if ((P \neq Q \land \neg x \underset{˙}{\leq} W), (ι z \in B \| \forall s \in A ((\neg s \underset{˙}{\leq} W \land s \underset{˙}{\lor} (x \underset{˙}{\land} W) = x) \to z = if (s \underset{˙}{\leq} (P \underset{˙}{\lor} Q), (ι y \in B \| \forall t \in A ((\neg t \underset{˙}{\leq} W \land \neg t \underset{˙}{\leq} (P \underset{˙}{\lor} Q)) \to y = E)), ⦋ s / t ⦌ D) \underset{˙}{\lor} (x \underset{˙}{\land} W))), x))$
	Assertion	cdlemg2dN	$⊢ ((K \in HL \land W \in H) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land (Q \in A \land \neg Q \underset{˙}{\leq} W)) \land (F \in T \land F (P) = Q)) \to F = G$

Proof

Step	Hyp	Ref	Expression
1		cdlemg2.b	$⊢ B = {Base}_{K}$
2		cdlemg2.l	$⊢ \underset{˙}{\leq} = \leq_{K}$
3		cdlemg2.j	$⊢ \underset{˙}{\lor} = join (K)$
4		cdlemg2.m	$⊢ \underset{˙}{\land} = meet (K)$
5		cdlemg2.a	$⊢ A = Atoms (K)$
6		cdlemg2.h	$⊢ H = LHyp (K)$
7		cdlemg2.t	$⊢ T = LTrn (K) (W)$
8		cdlemg2.u	$⊢ U = (P \underset{˙}{\lor} Q) \underset{˙}{\land} W$
9		cdlemg2.d	$⊢ D = (t \underset{˙}{\lor} U) \underset{˙}{\land} (Q \underset{˙}{\lor} ((P \underset{˙}{\lor} t) \underset{˙}{\land} W))$
10		cdlemg2.e	$⊢ E = (P \underset{˙}{\lor} Q) \underset{˙}{\land} (D \underset{˙}{\lor} ((s \underset{˙}{\lor} t) \underset{˙}{\land} W))$
11		cdlemg2.g	$⊢ G = (x \in B ⟼ if ((P \neq Q \land \neg x \underset{˙}{\leq} W), (ι z \in B \| \forall s \in A ((\neg s \underset{˙}{\leq} W \land s \underset{˙}{\lor} (x \underset{˙}{\land} W) = x) \to z = if (s \underset{˙}{\leq} (P \underset{˙}{\lor} Q), (ι y \in B \| \forall t \in A ((\neg t \underset{˙}{\leq} W \land \neg t \underset{˙}{\leq} (P \underset{˙}{\lor} Q)) \to y = E)), ⦋ s / t ⦌ D) \underset{˙}{\lor} (x \underset{˙}{\land} W))), x))$
12		simp3l	$⊢ ((K \in HL \land W \in H) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land (Q \in A \land \neg Q \underset{˙}{\leq} W)) \land (F \in T \land F (P) = Q)) \to F \in T$
13		simp1	$⊢ ((K \in HL \land W \in H) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land (Q \in A \land \neg Q \underset{˙}{\leq} W)) \land (F \in T \land F (P) = Q)) \to (K \in HL \land W \in H)$
14		simp2l	$⊢ ((K \in HL \land W \in H) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land (Q \in A \land \neg Q \underset{˙}{\leq} W)) \land (F \in T \land F (P) = Q)) \to (P \in A \land \neg P \underset{˙}{\leq} W)$
15		simp2r	$⊢ ((K \in HL \land W \in H) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land (Q \in A \land \neg Q \underset{˙}{\leq} W)) \land (F \in T \land F (P) = Q)) \to (Q \in A \land \neg Q \underset{˙}{\leq} W)$
16		simp3r	$⊢ ((K \in HL \land W \in H) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land (Q \in A \land \neg Q \underset{˙}{\leq} W)) \land (F \in T \land F (P) = Q)) \to F (P) = Q$
17	1 2 3 4 5 6 7 8 9 10 11	cdlemg2cN	$⊢ (((K \in HL \land W \in H) \land (P \in A \land \neg P \underset{˙}{\leq} W) \land (Q \in A \land \neg Q \underset{˙}{\leq} W)) \land F (P) = Q) \to (F \in T \leftrightarrow F = G)$
18	13 14 15 16 17	syl31anc	$⊢ ((K \in HL \land W \in H) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land (Q \in A \land \neg Q \underset{˙}{\leq} W)) \land (F \in T \land F (P) = Q)) \to (F \in T \leftrightarrow F = G)$
19	12 18	mpbid	$⊢ ((K \in HL \land W \in H) \land ((P \in A \land \neg P \underset{˙}{\leq} W) \land (Q \in A \land \neg Q \underset{˙}{\leq} W)) \land (F \in T \land F (P) = Q)) \to F = G$