Metamath Proof Explorer


Theorem nodenselem7

Description: Lemma for nodense . A and B are equal at all elements of the abstraction. (Contributed by Scott Fenton, 17-Jun-2011)

Ref Expression
Assertion nodenselem7 ( ( ( 𝐴 No 𝐵 No ) ∧ ( ( bday 𝐴 ) = ( bday 𝐵 ) ∧ 𝐴 <s 𝐵 ) ) → ( 𝐶 { 𝑎 ∈ On ∣ ( 𝐴𝑎 ) ≠ ( 𝐵𝑎 ) } → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 simpll ( ( ( 𝐴 No 𝐵 No ) ∧ ( ( bday 𝐴 ) = ( bday 𝐵 ) ∧ 𝐴 <s 𝐵 ) ) → 𝐴 No )
2 simplr ( ( ( 𝐴 No 𝐵 No ) ∧ ( ( bday 𝐴 ) = ( bday 𝐵 ) ∧ 𝐴 <s 𝐵 ) ) → 𝐵 No )
3 sltso <s Or No
4 sonr ( ( <s Or No 𝐴 No ) → ¬ 𝐴 <s 𝐴 )
5 3 4 mpan ( 𝐴 No → ¬ 𝐴 <s 𝐴 )
6 breq2 ( 𝐴 = 𝐵 → ( 𝐴 <s 𝐴𝐴 <s 𝐵 ) )
7 6 notbid ( 𝐴 = 𝐵 → ( ¬ 𝐴 <s 𝐴 ↔ ¬ 𝐴 <s 𝐵 ) )
8 5 7 syl5ibcom ( 𝐴 No → ( 𝐴 = 𝐵 → ¬ 𝐴 <s 𝐵 ) )
9 8 necon2ad ( 𝐴 No → ( 𝐴 <s 𝐵𝐴𝐵 ) )
10 9 imp ( ( 𝐴 No 𝐴 <s 𝐵 ) → 𝐴𝐵 )
11 10 ad2ant2rl ( ( ( 𝐴 No 𝐵 No ) ∧ ( ( bday 𝐴 ) = ( bday 𝐵 ) ∧ 𝐴 <s 𝐵 ) ) → 𝐴𝐵 )
12 1 2 11 3jca ( ( ( 𝐴 No 𝐵 No ) ∧ ( ( bday 𝐴 ) = ( bday 𝐵 ) ∧ 𝐴 <s 𝐵 ) ) → ( 𝐴 No 𝐵 No 𝐴𝐵 ) )
13 nosepeq ( ( ( 𝐴 No 𝐵 No 𝐴𝐵 ) ∧ 𝐶 { 𝑎 ∈ On ∣ ( 𝐴𝑎 ) ≠ ( 𝐵𝑎 ) } ) → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )
14 12 13 sylan ( ( ( ( 𝐴 No 𝐵 No ) ∧ ( ( bday 𝐴 ) = ( bday 𝐵 ) ∧ 𝐴 <s 𝐵 ) ) ∧ 𝐶 { 𝑎 ∈ On ∣ ( 𝐴𝑎 ) ≠ ( 𝐵𝑎 ) } ) → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )
15 14 ex ( ( ( 𝐴 No 𝐵 No ) ∧ ( ( bday 𝐴 ) = ( bday 𝐵 ) ∧ 𝐴 <s 𝐵 ) ) → ( 𝐶 { 𝑎 ∈ On ∣ ( 𝐴𝑎 ) ≠ ( 𝐵𝑎 ) } → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ) )