Metamath Proof Explorer


Theorem frirr

Description: A well-founded relation is irreflexive. Special case of Proposition 6.23 of TakeutiZaring p. 30. (Contributed by NM, 2-Jan-1994) (Revised by Mario Carneiro, 22-Jun-2015)

Ref Expression
Assertion frirr ( ( 𝑅 Fr 𝐴𝐵𝐴 ) → ¬ 𝐵 𝑅 𝐵 )

Proof

Step Hyp Ref Expression
1 simpl ( ( 𝑅 Fr 𝐴𝐵𝐴 ) → 𝑅 Fr 𝐴 )
2 snssi ( 𝐵𝐴 → { 𝐵 } ⊆ 𝐴 )
3 2 adantl ( ( 𝑅 Fr 𝐴𝐵𝐴 ) → { 𝐵 } ⊆ 𝐴 )
4 snnzg ( 𝐵𝐴 → { 𝐵 } ≠ ∅ )
5 4 adantl ( ( 𝑅 Fr 𝐴𝐵𝐴 ) → { 𝐵 } ≠ ∅ )
6 snex { 𝐵 } ∈ V
7 6 frc ( ( 𝑅 Fr 𝐴 ∧ { 𝐵 } ⊆ 𝐴 ∧ { 𝐵 } ≠ ∅ ) → ∃ 𝑦 ∈ { 𝐵 } { 𝑥 ∈ { 𝐵 } ∣ 𝑥 𝑅 𝑦 } = ∅ )
8 1 3 5 7 syl3anc ( ( 𝑅 Fr 𝐴𝐵𝐴 ) → ∃ 𝑦 ∈ { 𝐵 } { 𝑥 ∈ { 𝐵 } ∣ 𝑥 𝑅 𝑦 } = ∅ )
9 rabeq0 ( { 𝑥 ∈ { 𝐵 } ∣ 𝑥 𝑅 𝑦 } = ∅ ↔ ∀ 𝑥 ∈ { 𝐵 } ¬ 𝑥 𝑅 𝑦 )
10 breq2 ( 𝑦 = 𝐵 → ( 𝑥 𝑅 𝑦𝑥 𝑅 𝐵 ) )
11 10 notbid ( 𝑦 = 𝐵 → ( ¬ 𝑥 𝑅 𝑦 ↔ ¬ 𝑥 𝑅 𝐵 ) )
12 11 ralbidv ( 𝑦 = 𝐵 → ( ∀ 𝑥 ∈ { 𝐵 } ¬ 𝑥 𝑅 𝑦 ↔ ∀ 𝑥 ∈ { 𝐵 } ¬ 𝑥 𝑅 𝐵 ) )
13 9 12 syl5bb ( 𝑦 = 𝐵 → ( { 𝑥 ∈ { 𝐵 } ∣ 𝑥 𝑅 𝑦 } = ∅ ↔ ∀ 𝑥 ∈ { 𝐵 } ¬ 𝑥 𝑅 𝐵 ) )
14 13 rexsng ( 𝐵𝐴 → ( ∃ 𝑦 ∈ { 𝐵 } { 𝑥 ∈ { 𝐵 } ∣ 𝑥 𝑅 𝑦 } = ∅ ↔ ∀ 𝑥 ∈ { 𝐵 } ¬ 𝑥 𝑅 𝐵 ) )
15 breq1 ( 𝑥 = 𝐵 → ( 𝑥 𝑅 𝐵𝐵 𝑅 𝐵 ) )
16 15 notbid ( 𝑥 = 𝐵 → ( ¬ 𝑥 𝑅 𝐵 ↔ ¬ 𝐵 𝑅 𝐵 ) )
17 16 ralsng ( 𝐵𝐴 → ( ∀ 𝑥 ∈ { 𝐵 } ¬ 𝑥 𝑅 𝐵 ↔ ¬ 𝐵 𝑅 𝐵 ) )
18 14 17 bitrd ( 𝐵𝐴 → ( ∃ 𝑦 ∈ { 𝐵 } { 𝑥 ∈ { 𝐵 } ∣ 𝑥 𝑅 𝑦 } = ∅ ↔ ¬ 𝐵 𝑅 𝐵 ) )
19 18 adantl ( ( 𝑅 Fr 𝐴𝐵𝐴 ) → ( ∃ 𝑦 ∈ { 𝐵 } { 𝑥 ∈ { 𝐵 } ∣ 𝑥 𝑅 𝑦 } = ∅ ↔ ¬ 𝐵 𝑅 𝐵 ) )
20 8 19 mpbid ( ( 𝑅 Fr 𝐴𝐵𝐴 ) → ¬ 𝐵 𝑅 𝐵 )