Theorem precsexlem1 28620

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 6906	. 2 ⊢ (𝐿‘∅) = ((1st ∘ 𝐹)‘∅)
3		rdgfnon 8451	. . . 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 6655	. . . 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 6433	. . 3 ⊢ ∅ ∈ On
8		fvco2 7002	. . 3 ⊢ ((𝐹 Fn On ∧ ∅ ∈ On) → ((1st ∘ 𝐹)‘∅) = (1st ‘(𝐹‘∅)))
9	6, 7, 8	mp2an 705	. 2 ⊢ ((1st ∘ 𝐹)‘∅) = (1st ‘(𝐹‘∅))
10	4	fveq1i 6906	. . . . 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 5436	. . . . . 6 ⊢ ⟨{ 0s }, ∅⟩ ∈ V
12	11	rdg0 8454	. . . . 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 2786	. . . 4 ⊢ (𝐹‘∅) = ⟨{ 0s }, ∅⟩
14	13	fveq2i 6908	. . 3 ⊢ (1st ‘(𝐹‘∅)) = (1st ‘⟨{ 0s }, ∅⟩)
15		snex 5401	. . . 4 ⊢ { 0s } ∈ V
16		0ex 5264	. . . 4 ⊢ ∅ ∈ V
17	15, 16	op1st 8030	. . 3 ⊢ (1st ‘⟨{ 0s }, ∅⟩) = { 0s }
18	14, 17	eqtri 2786	. 2 ⊢ (1st ‘(𝐹‘∅)) = { 0s }
19	2, 9, 18	3eqtri 2790	1 ⊢ (𝐿‘∅) = { 0s }

This proof depends on syntax axioms:   = wceq 1570   ∈ wcel 2147  {cab 2741  ∃wrex 3090  {crab 3416  Vcvv 3454  ⦋csb 3851   ∪ cun 4152  ∅c0 4281  {csn 4588  ⟨cop 4838   class class class wbr 5106   ↦ cmpt 5189   ∘ ccom 5659  Oncon0 6377   Fn wfn 6550  ‘cfv 6555  (class class class)co 7440  1^st c1st 8020  2^nd c2nd 8021  reccrdg 8442   <s clts 28025   0_s c0s 28218   1_s c1s 28219  Lcleft 28238  Rcright 28239   +_s cadds 28372   −_s csubs 28433   ·_s cmuls 28519   /_su cdivs 28600

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-rep 5235  ax-sep 5252  ax-nul 5263  ax-pr 5395  ax-un 7772

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 2102  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3081  df-rex 3091  df-reu 3370  df-rab 3417  df-v 3456  df-sbc 3743  df-csb 3852  df-ss 3901  df-pss 4017  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-pw 4562  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-iun 4956  df-br 5107  df-opab 5171  df-mpt 5190  df-tr 5216  df-id 5550  df-eprel 5555  df-po 5563  df-so 5564  df-fr 5608  df-we 5610  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-ima 5668  df-pred 6319  df-ord 6380  df-on 6381  df-lim 6382  df-suc 6383  df-iota 6509  df-fun 6557  df-fn 6558  df-f 6559  df-f1 6560  df-fo 6561  df-f1o 6562  df-fv 6563  df-ov 7443  df-om 7899  df-1st 8022  df-2nd 8023  df-frecs 8324  df-wrecs 8355  df-recs 8404  df-rdg 8443

This theorem is used by:  precsexlem8  28627  precsexlem9  28628  precsexlem11  28630