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Metamath Proof Explorer

Theorem 2wlkdlem9

Description: Lemma 9 for 2wlkd . (Contributed by AV, 14-Feb-2021)

Ref Expression
Hypotheses 2wlkd.p ⊢ 𝑃 = ⟨“ 𝐴 𝐵 𝐶 ”⟩
2wlkd.f ⊢ 𝐹 = ⟨“ 𝐽 𝐾 ”⟩
2wlkd.s ⊢ ( 𝜑 → ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑉 ) )
2wlkd.n ⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 ∧ 𝐵 ≠ 𝐶 ) )
2wlkd.e ⊢ ( 𝜑 → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ 𝐽 ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ 𝐾 ) ) )
Assertion 2wlkdlem9 ( 𝜑 → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) ) )

Proof

Step Hyp Ref Expression
1 2wlkd.p ⊢ 𝑃 = ⟨“ 𝐴 𝐵 𝐶 ”⟩
2 2wlkd.f ⊢ 𝐹 = ⟨“ 𝐽 𝐾 ”⟩
3 2wlkd.s ⊢ ( 𝜑 → ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑉 ) )
4 2wlkd.n ⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 ∧ 𝐵 ≠ 𝐶 ) )
5 2wlkd.e ⊢ ( 𝜑 → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ 𝐽 ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ 𝐾 ) ) )
6 1 2 3 4 5 2wlkdlem8 ⊢ ( 𝜑 → ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) )
7 fveq2 ⊢ ( ( 𝐹 ‘ 0 ) = 𝐽 → ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) = ( 𝐼 ‘ 𝐽 ) )
8 7 adantr ⊢ ( ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) → ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) = ( 𝐼 ‘ 𝐽 ) )
9 8 sseq2d ⊢ ( ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) ↔ { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ 𝐽 ) ) )
10 fveq2 ⊢ ( ( 𝐹 ‘ 1 ) = 𝐾 → ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) = ( 𝐼 ‘ 𝐾 ) )
11 10 adantl ⊢ ( ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) → ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) = ( 𝐼 ‘ 𝐾 ) )
12 11 sseq2d ⊢ ( ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) → ( { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) ↔ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ 𝐾 ) ) )
13 9 12 anbi12d ⊢ ( ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) → ( ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) ) ↔ ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ 𝐽 ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ 𝐾 ) ) ) )
14 6 13 syl ⊢ ( 𝜑 → ( ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) ) ↔ ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ 𝐽 ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ 𝐾 ) ) ) )
15 5 14 mpbird ⊢ ( 𝜑 → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) ) )