Metamath Proof Explorer


Theorem n0lpligALT

Description: Alternate version of n0lplig using the predicate e/ instead of -. e. and whose proof bypasses nsnlplig . (Contributed by AV, 28-Nov-2021) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion n0lpligALT ( 𝐺 ∈ Plig → ∅ ∉ 𝐺 )

Proof

Step Hyp Ref Expression
1 eqid ⊢ ∪ 𝐺 = ∪ 𝐺
2 1 l2p ⊢ ( ( 𝐺 ∈ Plig ∧ ∅ ∈ 𝐺 ) → ∃ 𝑎 ∈ ∪ 𝐺 ∃ 𝑏 ∈ ∪ 𝐺 ( 𝑎 ≠ 𝑏 ∧ 𝑎 ∈ ∅ ∧ 𝑏 ∈ ∅ ) )
3 noel ⊢ ¬ 𝑎 ∈ ∅
4 3 pm2.21i ⊢ ( 𝑎 ∈ ∅ → ∅ ∉ 𝐺 )
5 4 3ad2ant2 ⊢ ( ( 𝑎 ≠ 𝑏 ∧ 𝑎 ∈ ∅ ∧ 𝑏 ∈ ∅ ) → ∅ ∉ 𝐺 )
6 5 a1i ⊢ ( ( 𝑎 ∈ ∪ 𝐺 ∧ 𝑏 ∈ ∪ 𝐺 ) → ( ( 𝑎 ≠ 𝑏 ∧ 𝑎 ∈ ∅ ∧ 𝑏 ∈ ∅ ) → ∅ ∉ 𝐺 ) )
7 6 rexlimivv ⊢ ( ∃ 𝑎 ∈ ∪ 𝐺 ∃ 𝑏 ∈ ∪ 𝐺 ( 𝑎 ≠ 𝑏 ∧ 𝑎 ∈ ∅ ∧ 𝑏 ∈ ∅ ) → ∅ ∉ 𝐺 )
8 2 7 syl ⊢ ( ( 𝐺 ∈ Plig ∧ ∅ ∈ 𝐺 ) → ∅ ∉ 𝐺 )
9 simpr ⊢ ( ( 𝐺 ∈ Plig ∧ ∅ ∉ 𝐺 ) → ∅ ∉ 𝐺 )
10 8 9 pm2.61danel ⊢ ( 𝐺 ∈ Plig → ∅ ∉ 𝐺 )