Metamath Proof Explorer


Theorem inagne3

Description: Deduce inequality from the in-angle relation. (Contributed by Thierry Arnoux, 29-Oct-2021)

Ref Expression
Hypotheses isinag.p ⊢ 𝑃 = ( Base ‘ 𝐺 )
isinag.i ⊢ 𝐼 = ( Itv ‘ 𝐺 )
isinag.k ⊢ 𝐾 = ( hlG ‘ 𝐺 )
isinag.x ⊢ ( 𝜑 → 𝑋 ∈ 𝑃 )
isinag.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
isinag.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
isinag.c ⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
inagflat.g ⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
inagswap.1 ⊢ ( 𝜑 → 𝑋 ( inA ‘ 𝐺 ) ⟨“ 𝐴 𝐵 𝐶 ”⟩ )
Assertion inagne3 ( 𝜑 → 𝑋 ≠ 𝐵 )

Proof

Step Hyp Ref Expression
1 isinag.p ⊢ 𝑃 = ( Base ‘ 𝐺 )
2 isinag.i ⊢ 𝐼 = ( Itv ‘ 𝐺 )
3 isinag.k ⊢ 𝐾 = ( hlG ‘ 𝐺 )
4 isinag.x ⊢ ( 𝜑 → 𝑋 ∈ 𝑃 )
5 isinag.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
6 isinag.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
7 isinag.c ⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
8 inagflat.g ⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
9 inagswap.1 ⊢ ( 𝜑 → 𝑋 ( inA ‘ 𝐺 ) ⟨“ 𝐴 𝐵 𝐶 ”⟩ )
10 1 2 3 4 5 6 7 8 isinag ⊢ ( 𝜑 → ( 𝑋 ( inA ‘ 𝐺 ) ⟨“ 𝐴 𝐵 𝐶 ”⟩ ↔ ( ( 𝐴 ≠ 𝐵 ∧ 𝐶 ≠ 𝐵 ∧ 𝑋 ≠ 𝐵 ) ∧ ∃ 𝑥 ∈ 𝑃 ( 𝑥 ∈ ( 𝐴 𝐼 𝐶 ) ∧ ( 𝑥 = 𝐵 ∨ 𝑥 ( 𝐾 ‘ 𝐵 ) 𝑋 ) ) ) ) )
11 9 10 mpbid ⊢ ( 𝜑 → ( ( 𝐴 ≠ 𝐵 ∧ 𝐶 ≠ 𝐵 ∧ 𝑋 ≠ 𝐵 ) ∧ ∃ 𝑥 ∈ 𝑃 ( 𝑥 ∈ ( 𝐴 𝐼 𝐶 ) ∧ ( 𝑥 = 𝐵 ∨ 𝑥 ( 𝐾 ‘ 𝐵 ) 𝑋 ) ) ) )
12 11 simpld ⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 ∧ 𝐶 ≠ 𝐵 ∧ 𝑋 ≠ 𝐵 ) )
13 12 simp3d ⊢ ( 𝜑 → 𝑋 ≠ 𝐵 )