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 → ( 𝐴𝑉𝐵𝑉𝐶𝑉 ) )
2wlkd.n ( 𝜑 → ( 𝐴𝐵𝐵𝐶 ) )
2wlkd.e ( 𝜑 → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐽 ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼𝐾 ) ) )
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 → ( 𝐴𝑉𝐵𝑉𝐶𝑉 ) )
4 2wlkd.n ( 𝜑 → ( 𝐴𝐵𝐵𝐶 ) )
5 2wlkd.e ( 𝜑 → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐽 ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼𝐾 ) ) )
6 1 2 3 4 5 2wlkdlem8 ( 𝜑 → ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) )
7 fveq2 ( ( 𝐹 ‘ 0 ) = 𝐽 → ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) = ( 𝐼𝐽 ) )
8 7 adantr ( ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) → ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) = ( 𝐼𝐽 ) )
9 8 sseq2d ( ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) ↔ { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐽 ) ) )
10 fveq2 ( ( 𝐹 ‘ 1 ) = 𝐾 → ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) = ( 𝐼𝐾 ) )
11 10 adantl ( ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) → ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) = ( 𝐼𝐾 ) )
12 11 sseq2d ( ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) → ( { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) ↔ { 𝐵 , 𝐶 } ⊆ ( 𝐼𝐾 ) ) )
13 9 12 anbi12d ( ( ( 𝐹 ‘ 0 ) = 𝐽 ∧ ( 𝐹 ‘ 1 ) = 𝐾 ) → ( ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) ) ↔ ( { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐽 ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼𝐾 ) ) ) )
14 6 13 syl ( 𝜑 → ( ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) ) ↔ ( { 𝐴 , 𝐵 } ⊆ ( 𝐼𝐽 ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼𝐾 ) ) ) )
15 5 14 mpbird ( 𝜑 → ( { 𝐴 , 𝐵 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 0 ) ) ∧ { 𝐵 , 𝐶 } ⊆ ( 𝐼 ‘ ( 𝐹 ‘ 1 ) ) ) )