Theorem precsexlem1 28470

Description: Lemma for surreal reciprocals. Calculate the value of the recursive left function at zero. (Contributed by Scott Fenton, 13-Mar-2025.)

Hypotheses

Ref	Expression
precsexlem.1	⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)
precsexlem.2	⊢ 𝐿 = (1st ∘ 𝐹)
precsexlem.3	⊢ 𝑅 = (2nd ∘ 𝐹)

Assertion

Ref	Expression
precsexlem1	⊢ (𝐿‘∅) = { 0s }

Proof of Theorem precsexlem1

Step	Hyp	Ref	Expression
1		precsexlem.2	. . 3 ⊢ 𝐿 = (1st ∘ 𝐹)
2	1	fveq1i 6883	. 2 ⊢ (𝐿‘∅) = ((1st ∘ 𝐹)‘∅)
3		rdgfnon 8410	. . . 4 ⊢ rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩) Fn On
4		precsexlem.1	. . . . 5 ⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)
5	4	fneq1i 6633	. . . 4 ⊢ (𝐹 Fn On ↔ rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩) Fn On)
6	3, 5	mpbir 234	. . 3 ⊢ 𝐹 Fn On
7		0elon 6417	. . 3 ⊢ ∅ ∈ On
8		fvco2 6979	. . 3 ⊢ ((𝐹 Fn On ∧ ∅ ∈ On) → ((1st ∘ 𝐹)‘∅) = (1st ‘(𝐹‘∅)))
9	6, 7, 8	mp2an 705	. 2 ⊢ ((1st ∘ 𝐹)‘∅) = (1st ‘(𝐹‘∅))
10	4	fveq1i 6883	. . . . 5 ⊢ (𝐹‘∅) = (rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)‘∅)
11		opex 5443	. . . . . 6 ⊢ ⟨{ 0s }, ∅⟩ ∈ V
12	11	rdg0 8413	. . . . 5 ⊢ (rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)‘∅) = ⟨{ 0s }, ∅⟩
13	10, 12	eqtri 2785	. . . 4 ⊢ (𝐹‘∅) = ⟨{ 0s }, ∅⟩
14	13	fveq2i 6885	. . 3 ⊢ (1st ‘(𝐹‘∅)) = (1st ‘⟨{ 0s }, ∅⟩)
15		snex 5408	. . . 4 ⊢ { 0s } ∈ V
16		0ex 5268	. . . 4 ⊢ ∅ ∈ V
17	15, 16	op1st 7997	. . 3 ⊢ (1st ‘⟨{ 0s }, ∅⟩) = { 0s }
18	14, 17	eqtri 2785	. 2 ⊢ (1st ‘(𝐹‘∅)) = { 0s }
19	2, 9, 18	3eqtri 2789	1 ⊢ (𝐿‘∅) = { 0s }

This proof depends on syntax axioms:   = wceq 1570   ∈ wcel 2145  {cab 2740  ∃wrex 3088  {crab 3414  Vcvv 3453  ⦋csb 3850   ∪ cun 3900  ∅c0 4282  {csn 4587  ⟨cop 4593   class class class wbr 5107   ↦ cmpt 5190   ∘ ccom 5663  Oncon0 6361   Fn wfn 6532  ‘cfv 6537  (class class class)co 7416  1^st c1st 7987  2^nd c2nd 7988  reccrdg 8401   <s clts 27875   0_s c0s 28068   1_s c1s 28069   L cleft 28088   R cright 28089   +_s cadds 28222   -_s csubs 28283   ·_s cmuls 28369   /_su cdivs 28450

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734  ax-rep 5236  ax-sep 5255  ax-nul 5267  ax-pr 5402  ax-un 7739

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3or 1104  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ne 2958  df-ral 3079  df-rex 3089  df-reu 3368  df-rab 3415  df-v 3455  df-sbc 3743  df-csb 3851  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-pss 3922  df-nul 4283  df-if 4486  df-pw 4562  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-iun 4956  df-br 5108  df-opab 5172  df-mpt 5191  df-tr 5217  df-id 5554  df-eprel 5559  df-po 5567  df-so 5568  df-fr 5612  df-we 5614  df-xp 5665  df-rel 5666  df-cnv 5667  df-co 5668  df-dm 5669  df-rn 5670  df-res 5671  df-ima 5672  df-pred 6303  df-ord 6364  df-on 6365  df-lim 6366  df-suc 6367  df-iota 6493  df-fun 6539  df-fn 6540  df-f 6541  df-f1 6542  df-fo 6543  df-f1o 6544  df-fv 6545  df-ov 7419  df-om 7866  df-1st 7989  df-2nd 7990  df-frecs 8283  df-wrecs 8314  df-recs 8363  df-rdg 8402

This theorem is used by:  precsexlem8  28477  precsexlem9  28478  precsexlem11  28480