findsg

Description: Principle of Finite Induction (inference schema), using implicit substitutions. The first four hypotheses establish the substitutions we need. The last two are the basis and the induction step. The basis of this version is an arbitrary natural number B instead of zero. (Contributed by NM, 16-Sep-1995)

	Ref	Expression
Hypotheses	findsg.1	$⊢ x = B \to (φ \leftrightarrow ψ)$
	findsg.2	$⊢ x = y \to (φ \leftrightarrow χ)$
	findsg.3	$⊢ x = suc y \to (φ \leftrightarrow θ)$
	findsg.4	$⊢ x = A \to (φ \leftrightarrow τ)$
	findsg.5	$⊢ B \in ω \to ψ$
	findsg.6	$⊢ ((y \in ω \land B \in ω) \land B \subseteq y) \to (χ \to θ)$
Assertion	findsg	$⊢ ((A \in ω \land B \in ω) \land B \subseteq A) \to τ$

Proof

Step	Hyp	Ref	Expression
1		findsg.1	$⊢ x = B \to (φ \leftrightarrow ψ)$
2		findsg.2	$⊢ x = y \to (φ \leftrightarrow χ)$
3		findsg.3	$⊢ x = suc y \to (φ \leftrightarrow θ)$
4		findsg.4	$⊢ x = A \to (φ \leftrightarrow τ)$
5		findsg.5	$⊢ B \in ω \to ψ$
6		findsg.6	$⊢ ((y \in ω \land B \in ω) \land B \subseteq y) \to (χ \to θ)$
7		sseq2	$⊢ x = \emptyset \to (B \subseteq x \leftrightarrow B \subseteq \emptyset)$
8	7	adantl	$⊢ (B = \emptyset \land x = \emptyset) \to (B \subseteq x \leftrightarrow B \subseteq \emptyset)$
9		eqeq2	$⊢ B = \emptyset \to (x = B \leftrightarrow x = \emptyset)$
10	9 1	syl6bir	$⊢ B = \emptyset \to (x = \emptyset \to (φ \leftrightarrow ψ))$
11	10	imp	$⊢ (B = \emptyset \land x = \emptyset) \to (φ \leftrightarrow ψ)$
12	8 11	imbi12d	$⊢ (B = \emptyset \land x = \emptyset) \to ((B \subseteq x \to φ) \leftrightarrow (B \subseteq \emptyset \to ψ))$
13	7	imbi1d	$⊢ x = \emptyset \to ((B \subseteq x \to φ) \leftrightarrow (B \subseteq \emptyset \to φ))$
14		ss0	$⊢ B \subseteq \emptyset \to B = \emptyset$
15	14	con3i	$⊢ \neg B = \emptyset \to \neg B \subseteq \emptyset$
16	15	pm2.21d	$⊢ \neg B = \emptyset \to (B \subseteq \emptyset \to (φ \leftrightarrow ψ))$
17	16	pm5.74d	$⊢ \neg B = \emptyset \to ((B \subseteq \emptyset \to φ) \leftrightarrow (B \subseteq \emptyset \to ψ))$
18	13 17	sylan9bbr	$⊢ (\neg B = \emptyset \land x = \emptyset) \to ((B \subseteq x \to φ) \leftrightarrow (B \subseteq \emptyset \to ψ))$
19	12 18	pm2.61ian	$⊢ x = \emptyset \to ((B \subseteq x \to φ) \leftrightarrow (B \subseteq \emptyset \to ψ))$
20	19	imbi2d	$⊢ x = \emptyset \to ((B \in ω \to (B \subseteq x \to φ)) \leftrightarrow (B \in ω \to (B \subseteq \emptyset \to ψ)))$
21		sseq2	$⊢ x = y \to (B \subseteq x \leftrightarrow B \subseteq y)$
22	21 2	imbi12d	$⊢ x = y \to ((B \subseteq x \to φ) \leftrightarrow (B \subseteq y \to χ))$
23	22	imbi2d	$⊢ x = y \to ((B \in ω \to (B \subseteq x \to φ)) \leftrightarrow (B \in ω \to (B \subseteq y \to χ)))$
24		sseq2	$⊢ x = suc y \to (B \subseteq x \leftrightarrow B \subseteq suc y)$
25	24 3	imbi12d	$⊢ x = suc y \to ((B \subseteq x \to φ) \leftrightarrow (B \subseteq suc y \to θ))$
26	25	imbi2d	$⊢ x = suc y \to ((B \in ω \to (B \subseteq x \to φ)) \leftrightarrow (B \in ω \to (B \subseteq suc y \to θ)))$
27		sseq2	$⊢ x = A \to (B \subseteq x \leftrightarrow B \subseteq A)$
28	27 4	imbi12d	$⊢ x = A \to ((B \subseteq x \to φ) \leftrightarrow (B \subseteq A \to τ))$
29	28	imbi2d	$⊢ x = A \to ((B \in ω \to (B \subseteq x \to φ)) \leftrightarrow (B \in ω \to (B \subseteq A \to τ)))$
30	5	a1d	$⊢ B \in ω \to (B \subseteq \emptyset \to ψ)$
31		vex	$⊢ y \in V$
32	31	sucex	$⊢ suc y \in V$
33	32	eqvinc	$⊢ suc y = B \leftrightarrow \exists x (x = suc y \land x = B)$
34	5 1	imbitrrid	$⊢ x = B \to (B \in ω \to φ)$
35	3	biimpd	$⊢ x = suc y \to (φ \to θ)$
36	34 35	sylan9r	$⊢ (x = suc y \land x = B) \to (B \in ω \to θ)$
37	36	exlimiv	$⊢ \exists x (x = suc y \land x = B) \to (B \in ω \to θ)$
38	33 37	sylbi	$⊢ suc y = B \to (B \in ω \to θ)$
39	38	eqcoms	$⊢ B = suc y \to (B \in ω \to θ)$
40	39	imim2i	$⊢ (B \subseteq suc y \to B = suc y) \to (B \subseteq suc y \to (B \in ω \to θ))$
41	40	a1d	$⊢ (B \subseteq suc y \to B = suc y) \to ((B \subseteq y \to χ) \to (B \subseteq suc y \to (B \in ω \to θ)))$
42	41	com4r	$⊢ B \in ω \to ((B \subseteq suc y \to B = suc y) \to ((B \subseteq y \to χ) \to (B \subseteq suc y \to θ)))$
43	42	adantl	$⊢ (y \in ω \land B \in ω) \to ((B \subseteq suc y \to B = suc y) \to ((B \subseteq y \to χ) \to (B \subseteq suc y \to θ)))$
44		df-ne	$⊢ B \neq suc y \leftrightarrow \neg B = suc y$
45	44	anbi2i	$⊢ (B \subseteq suc y \land B \neq suc y) \leftrightarrow (B \subseteq suc y \land \neg B = suc y)$
46		annim	$⊢ (B \subseteq suc y \land \neg B = suc y) \leftrightarrow \neg (B \subseteq suc y \to B = suc y)$
47	45 46	bitri	$⊢ (B \subseteq suc y \land B \neq suc y) \leftrightarrow \neg (B \subseteq suc y \to B = suc y)$
48		nnon	$⊢ B \in ω \to B \in On$
49		nnon	$⊢ y \in ω \to y \in On$
50		onsssuc	$⊢ (B \in On \land y \in On) \to (B \subseteq y \leftrightarrow B \in suc y)$
51		onsuc	$⊢ y \in On \to suc y \in On$
52		onelpss	$⊢ (B \in On \land suc y \in On) \to (B \in suc y \leftrightarrow (B \subseteq suc y \land B \neq suc y))$
53	51 52	sylan2	$⊢ (B \in On \land y \in On) \to (B \in suc y \leftrightarrow (B \subseteq suc y \land B \neq suc y))$
54	50 53	bitrd	$⊢ (B \in On \land y \in On) \to (B \subseteq y \leftrightarrow (B \subseteq suc y \land B \neq suc y))$
55	48 49 54	syl2anr	$⊢ (y \in ω \land B \in ω) \to (B \subseteq y \leftrightarrow (B \subseteq suc y \land B \neq suc y))$
56	6	ex	$⊢ (y \in ω \land B \in ω) \to (B \subseteq y \to (χ \to θ))$
57	56	a1ddd	$⊢ (y \in ω \land B \in ω) \to (B \subseteq y \to (χ \to (B \subseteq suc y \to θ)))$
58	57	a2d	$⊢ (y \in ω \land B \in ω) \to ((B \subseteq y \to χ) \to (B \subseteq y \to (B \subseteq suc y \to θ)))$
59	58	com23	$⊢ (y \in ω \land B \in ω) \to (B \subseteq y \to ((B \subseteq y \to χ) \to (B \subseteq suc y \to θ)))$
60	55 59	sylbird	$⊢ (y \in ω \land B \in ω) \to ((B \subseteq suc y \land B \neq suc y) \to ((B \subseteq y \to χ) \to (B \subseteq suc y \to θ)))$
61	47 60	biimtrrid	$⊢ (y \in ω \land B \in ω) \to (\neg (B \subseteq suc y \to B = suc y) \to ((B \subseteq y \to χ) \to (B \subseteq suc y \to θ)))$
62	43 61	pm2.61d	$⊢ (y \in ω \land B \in ω) \to ((B \subseteq y \to χ) \to (B \subseteq suc y \to θ))$
63	62	ex	$⊢ y \in ω \to (B \in ω \to ((B \subseteq y \to χ) \to (B \subseteq suc y \to θ)))$
64	63	a2d	$⊢ y \in ω \to ((B \in ω \to (B \subseteq y \to χ)) \to (B \in ω \to (B \subseteq suc y \to θ)))$
65	20 23 26 29 30 64	finds	$⊢ A \in ω \to (B \in ω \to (B \subseteq A \to τ))$
66	65	imp31	$⊢ ((A \in ω \land B \in ω) \land B \subseteq A) \to τ$

Metamath Proof Explorer

Theorem findsg

Proof