Theorem addscutlem

Description: Lemma for addscut . Show the statement with some additional distinct variable conditions. (Contributed by Scott Fenton, 8-Mar-2025)

	Ref	Expression
Hypotheses	addscutlem.1	⊢ ( 𝜑 → 𝑋 ∈ No )
	addscutlem.2	⊢ ( 𝜑 → 𝑌 ∈ No )
Assertion	addscutlem	⊢ ( 𝜑 → ( ( 𝑋 +s 𝑌 ) ∈ No ∧ ( { 𝑝 ∣ ∃ 𝑙 ∈ ( L ‘ 𝑋 ) 𝑝 = ( 𝑙 +s 𝑌 ) } ∪ { 𝑞 ∣ ∃ 𝑚 ∈ ( L ‘ 𝑌 ) 𝑞 = ( 𝑋 +s 𝑚 ) } ) <<s { ( 𝑋 +s 𝑌 ) } ∧ { ( 𝑋 +s 𝑌 ) } <<s ( { 𝑤 ∣ ∃ 𝑟 ∈ ( R ‘ 𝑋 ) 𝑤 = ( 𝑟 +s 𝑌 ) } ∪ { 𝑡 ∣ ∃ 𝑠 ∈ ( R ‘ 𝑌 ) 𝑡 = ( 𝑋 +s 𝑠 ) } ) ) )

Proof

Step	Hyp	Ref	Expression
1		addscutlem.1	⊢ ( 𝜑 → 𝑋 ∈ No )
2		addscutlem.2	⊢ ( 𝜑 → 𝑌 ∈ No )
3		addsprop	⊢ ( ( 𝑥 ∈ No ∧ 𝑦 ∈ No ∧ 𝑧 ∈ No ) → ( ( 𝑥 +s 𝑦 ) ∈ No ∧ ( 𝑦 <s 𝑧 → ( 𝑦 +s 𝑥 ) <s ( 𝑧 +s 𝑥 ) ) ) )
4	3	a1d	⊢ ( ( 𝑥 ∈ No ∧ 𝑦 ∈ No ∧ 𝑧 ∈ No ) → ( ( ( ( bday ‘ 𝑥 ) +no ( bday ‘ 𝑦 ) ) ∪ ( ( bday ‘ 𝑥 ) +no ( bday ‘ 𝑧 ) ) ) ∈ ( ( ( bday ‘ 𝑋 ) +no ( bday ‘ 𝑌 ) ) ∪ ( ( bday ‘ 𝑋 ) +no ( bday ‘ 0s ) ) ) → ( ( 𝑥 +s 𝑦 ) ∈ No ∧ ( 𝑦 <s 𝑧 → ( 𝑦 +s 𝑥 ) <s ( 𝑧 +s 𝑥 ) ) ) ) )
5	4	rgen3	⊢ ∀ 𝑥 ∈ No ∀ 𝑦 ∈ No ∀ 𝑧 ∈ No ( ( ( ( bday ‘ 𝑥 ) +no ( bday ‘ 𝑦 ) ) ∪ ( ( bday ‘ 𝑥 ) +no ( bday ‘ 𝑧 ) ) ) ∈ ( ( ( bday ‘ 𝑋 ) +no ( bday ‘ 𝑌 ) ) ∪ ( ( bday ‘ 𝑋 ) +no ( bday ‘ 0s ) ) ) → ( ( 𝑥 +s 𝑦 ) ∈ No ∧ ( 𝑦 <s 𝑧 → ( 𝑦 +s 𝑥 ) <s ( 𝑧 +s 𝑥 ) ) ) )
6	5	a1i	⊢ ( 𝜑 → ∀ 𝑥 ∈ No ∀ 𝑦 ∈ No ∀ 𝑧 ∈ No ( ( ( ( bday ‘ 𝑥 ) +no ( bday ‘ 𝑦 ) ) ∪ ( ( bday ‘ 𝑥 ) +no ( bday ‘ 𝑧 ) ) ) ∈ ( ( ( bday ‘ 𝑋 ) +no ( bday ‘ 𝑌 ) ) ∪ ( ( bday ‘ 𝑋 ) +no ( bday ‘ 0s ) ) ) → ( ( 𝑥 +s 𝑦 ) ∈ No ∧ ( 𝑦 <s 𝑧 → ( 𝑦 +s 𝑥 ) <s ( 𝑧 +s 𝑥 ) ) ) ) )
7	6 1 2	addsproplem3	⊢ ( 𝜑 → ( ( 𝑋 +s 𝑌 ) ∈ No ∧ ( { 𝑝 ∣ ∃ 𝑙 ∈ ( L ‘ 𝑋 ) 𝑝 = ( 𝑙 +s 𝑌 ) } ∪ { 𝑞 ∣ ∃ 𝑚 ∈ ( L ‘ 𝑌 ) 𝑞 = ( 𝑋 +s 𝑚 ) } ) <<s { ( 𝑋 +s 𝑌 ) } ∧ { ( 𝑋 +s 𝑌 ) } <<s ( { 𝑤 ∣ ∃ 𝑟 ∈ ( R ‘ 𝑋 ) 𝑤 = ( 𝑟 +s 𝑌 ) } ∪ { 𝑡 ∣ ∃ 𝑠 ∈ ( R ‘ 𝑌 ) 𝑡 = ( 𝑋 +s 𝑠 ) } ) ) )