Theorem psubsuccv-P6-L1 805

Description: Lemma for psubsuccv-P6 806.

Hypothesis

Ref	Expression
psubsuccv-P6-L1.1	⊢ ([𝑤 / 𝑥] 𝑎 = 𝑡 ↔ 𝑎 = 𝑢)

Assertion

Ref	Expression
psubsuccv-P6-L1	⊢ (𝑥 = 𝑤 → (𝑏 = s‘𝑡 ↔ 𝑏 = s‘𝑢))

Distinct variable groups:   𝑡,𝑎   𝑢,𝑎   𝑤,𝑎   𝑡,𝑏   𝑢,𝑏   𝑤,𝑏,𝑎   𝑥,𝑎,𝑏

Proof of Theorem psubsuccv-P6-L1

Step	Hyp	Ref	Expression
1		rcp-NDASM1of2 193	. . . . . . 7 ⊢ ((𝑎 = 𝑡 ∧ 𝑥 = 𝑤) → 𝑎 = 𝑡)
2	1	eqsym-P5 627	. . . . . 6 ⊢ ((𝑎 = 𝑡 ∧ 𝑥 = 𝑤) → 𝑡 = 𝑎)
3		psubtoisub-P6 765	. . . . . . . . . 10 ⊢ (𝑥 = 𝑤 → (𝑎 = 𝑡 ↔ [𝑤 / 𝑥] 𝑎 = 𝑡))
4		psubsuccv-P6-L1.1	. . . . . . . . . . 11 ⊢ ([𝑤 / 𝑥] 𝑎 = 𝑡 ↔ 𝑎 = 𝑢)
5	4	rcp-NDIMP0addall 207	. . . . . . . . . 10 ⊢ (𝑥 = 𝑤 → ([𝑤 / 𝑥] 𝑎 = 𝑡 ↔ 𝑎 = 𝑢))
6	3, 5	bitrns-P3.33c 302	. . . . . . . . 9 ⊢ (𝑥 = 𝑤 → (𝑎 = 𝑡 ↔ 𝑎 = 𝑢))
7	6	rcp-NDIMP0addall 207	. . . . . . . 8 ⊢ (𝑎 = 𝑡 → (𝑥 = 𝑤 → (𝑎 = 𝑡 ↔ 𝑎 = 𝑢)))
8	7	import-P3.34a.RC 306	. . . . . . 7 ⊢ ((𝑎 = 𝑡 ∧ 𝑥 = 𝑤) → (𝑎 = 𝑡 ↔ 𝑎 = 𝑢))
9	1, 8	bimpf-P4 531	. . . . . 6 ⊢ ((𝑎 = 𝑡 ∧ 𝑥 = 𝑤) → 𝑎 = 𝑢)
10	2, 9	eqtrns-P5 630	. . . . 5 ⊢ ((𝑎 = 𝑡 ∧ 𝑥 = 𝑤) → 𝑡 = 𝑢)
11	10	subsucc-P5 644	. . . 4 ⊢ ((𝑎 = 𝑡 ∧ 𝑥 = 𝑤) → s‘𝑡 = s‘𝑢)
12	11	subeqr-P5 635	. . 3 ⊢ ((𝑎 = 𝑡 ∧ 𝑥 = 𝑤) → (𝑏 = s‘𝑡 ↔ 𝑏 = s‘𝑢))
13	12	rcp-NDIMI2 224	. 2 ⊢ (𝑎 = 𝑡 → (𝑥 = 𝑤 → (𝑏 = s‘𝑡 ↔ 𝑏 = s‘𝑢)))
14		axL6ex-P5 625	. 2 ⊢ ∃𝑎 𝑎 = 𝑡
15	13, 14	exiav-P5.SH 616	1 ⊢ (𝑥 = 𝑤 → (𝑏 = s‘𝑡 ↔ 𝑏 = s‘𝑢))

Syntax hints:  term-obj 1  s‘term_succ 3   = wff-equals 6   → wff-imp 10   ↔ wff-bi 104   ∧ wff-and 132  [wff-psub 714

This theorem was proved from axioms:  ax-L1 11  ax-L2 12  ax-L3 13  ax-MP 14  ax-GEN 15  ax-L4 16  ax-L5 17  ax-L6 18  ax-L7 19  ax-L9-succ 22  ax-L10 27  ax-L12 29

This theorem depends on definitions:  df-bi-D2.1 107  df-and-D2.2 133  df-or-D2.3 145  df-true-D2.4 155  df-rcp-AND3 161  df-exists-D5.1 596  df-nfree-D6.1 682  df-psub-D6.2 716

This theorem is referenced by:  psubsuccv-P6  806