Metamath Proof Explorer


Theorem chndin

Description: A chain in two alphabets at once is also a chain in their intersection. (Contributed by Ender Ting, 24-Jul-2026)

Ref Expression
Assertion chndin ( ( 𝐴 ∈ ( 𝑅 Chain 𝐵 ) ∧ 𝐴 ∈ ( 𝑅 Chain 𝐶 ) ) → 𝐴 ∈ ( 𝑅 Chain ( 𝐵𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 id ( 𝐴 ∈ ( 𝑅 Chain 𝐵 ) → 𝐴 ∈ ( 𝑅 Chain 𝐵 ) )
2 1 chnwrd ( 𝐴 ∈ ( 𝑅 Chain 𝐵 ) → 𝐴 ∈ Word 𝐵 )
3 id ( 𝐴 ∈ ( 𝑅 Chain 𝐶 ) → 𝐴 ∈ ( 𝑅 Chain 𝐶 ) )
4 3 chnwrd ( 𝐴 ∈ ( 𝑅 Chain 𝐶 ) → 𝐴 ∈ Word 𝐶 )
5 wrddin ( ( 𝐴 ∈ Word 𝐵𝐴 ∈ Word 𝐶 ) → 𝐴 ∈ Word ( 𝐵𝐶 ) )
6 2 4 5 syl2an ( ( 𝐴 ∈ ( 𝑅 Chain 𝐵 ) ∧ 𝐴 ∈ ( 𝑅 Chain 𝐶 ) ) → 𝐴 ∈ Word ( 𝐵𝐶 ) )
7 ischn ( 𝐴 ∈ ( 𝑅 Chain 𝐵 ) ↔ ( 𝐴 ∈ Word 𝐵 ∧ ∀ 𝑛 ∈ ( dom 𝐴 ∖ { 0 } ) ( 𝐴 ‘ ( 𝑛 − 1 ) ) 𝑅 ( 𝐴𝑛 ) ) )
8 7 simprbi ( 𝐴 ∈ ( 𝑅 Chain 𝐵 ) → ∀ 𝑛 ∈ ( dom 𝐴 ∖ { 0 } ) ( 𝐴 ‘ ( 𝑛 − 1 ) ) 𝑅 ( 𝐴𝑛 ) )
9 8 adantr ( ( 𝐴 ∈ ( 𝑅 Chain 𝐵 ) ∧ 𝐴 ∈ ( 𝑅 Chain 𝐶 ) ) → ∀ 𝑛 ∈ ( dom 𝐴 ∖ { 0 } ) ( 𝐴 ‘ ( 𝑛 − 1 ) ) 𝑅 ( 𝐴𝑛 ) )
10 ischn ( 𝐴 ∈ ( 𝑅 Chain ( 𝐵𝐶 ) ) ↔ ( 𝐴 ∈ Word ( 𝐵𝐶 ) ∧ ∀ 𝑛 ∈ ( dom 𝐴 ∖ { 0 } ) ( 𝐴 ‘ ( 𝑛 − 1 ) ) 𝑅 ( 𝐴𝑛 ) ) )
11 6 9 10 sylanbrc ( ( 𝐴 ∈ ( 𝑅 Chain 𝐵 ) ∧ 𝐴 ∈ ( 𝑅 Chain 𝐶 ) ) → 𝐴 ∈ ( 𝑅 Chain ( 𝐵𝐶 ) ) )