Theorem precsexlem2

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

	Ref	Expression
Hypotheses	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	precsexlem2	⊢ ( 𝑅 ‘ ∅ ) = ∅

Proof

Step	Hyp	Ref	Expression
1		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 } , ∅ ⟩ )
2		precsexlem.2	⊢ 𝐿 = ( 1st ∘ 𝐹 )
3		precsexlem.3	⊢ 𝑅 = ( 2nd ∘ 𝐹 )
4	3	fveq1i	⊢ ( 𝑅 ‘ ∅ ) = ( ( 2nd ∘ 𝐹 ) ‘ ∅ )
5		rdgfnon	⊢ 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	1	fneq1i	⊢ ( 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 )
7	5 6	mpbir	⊢ 𝐹 Fn On
8		0elon	⊢ ∅ ∈ On
9		fvco2	⊢ ( ( 𝐹 Fn On ∧ ∅ ∈ On ) → ( ( 2nd ∘ 𝐹 ) ‘ ∅ ) = ( 2nd ‘ ( 𝐹 ‘ ∅ ) ) )
10	7 8 9	mp2an	⊢ ( ( 2nd ∘ 𝐹 ) ‘ ∅ ) = ( 2nd ‘ ( 𝐹 ‘ ∅ ) )
11	1	fveq1i	⊢ ( 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 } , ∅ ⟩ ) ‘ ∅ )
12		opex	⊢ ⟨ { 0s } , ∅ ⟩ ∈ V
13	12	rdg0	⊢ ( 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 } , ∅ ⟩
14	11 13	eqtri	⊢ ( 𝐹 ‘ ∅ ) = ⟨ { 0s } , ∅ ⟩
15	14	fveq2i	⊢ ( 2nd ‘ ( 𝐹 ‘ ∅ ) ) = ( 2nd ‘ ⟨ { 0s } , ∅ ⟩ )
16		snex	⊢ { 0s } ∈ V
17		0ex	⊢ ∅ ∈ V
18	16 17	op2nd	⊢ ( 2nd ‘ ⟨ { 0s } , ∅ ⟩ ) = ∅
19	15 18	eqtri	⊢ ( 2nd ‘ ( 𝐹 ‘ ∅ ) ) = ∅
20	4 10 19	3eqtri	⊢ ( 𝑅 ‘ ∅ ) = ∅