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Description: Lemma 3 for clwlkclwwlklem2a . (Contributed by Alexander van der Vekens, 21-Jun-2018)
| Ref | Expression | ||
|---|---|---|---|
| Hypothesis | clwlkclwwlklem2.f | ⊢ 𝐹 = ( 𝑥 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑃 ) − 1 ) ) ↦ if ( 𝑥 < ( ( ♯ ‘ 𝑃 ) − 2 ) , ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ ( 𝑥 + 1 ) ) } ) , ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ 0 ) } ) ) ) | |
| Assertion | clwlkclwwlklem2a3 | ⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( 𝑃 ‘ ( ♯ ‘ 𝐹 ) ) = ( lastS ‘ 𝑃 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | clwlkclwwlklem2.f | ⊢ 𝐹 = ( 𝑥 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑃 ) − 1 ) ) ↦ if ( 𝑥 < ( ( ♯ ‘ 𝑃 ) − 2 ) , ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ ( 𝑥 + 1 ) ) } ) , ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ 0 ) } ) ) ) | |
| 2 | lsw | ⊢ ( 𝑃 ∈ Word 𝑉 → ( lastS ‘ 𝑃 ) = ( 𝑃 ‘ ( ( ♯ ‘ 𝑃 ) − 1 ) ) ) | |
| 3 | 2 | adantr | ⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( lastS ‘ 𝑃 ) = ( 𝑃 ‘ ( ( ♯ ‘ 𝑃 ) − 1 ) ) ) |
| 4 | 1 | clwlkclwwlklem2a2 | ⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) ) |
| 5 | 4 | eqcomd | ⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( ( ♯ ‘ 𝑃 ) − 1 ) = ( ♯ ‘ 𝐹 ) ) |
| 6 | 5 | fveq2d | ⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( 𝑃 ‘ ( ( ♯ ‘ 𝑃 ) − 1 ) ) = ( 𝑃 ‘ ( ♯ ‘ 𝐹 ) ) ) |
| 7 | 3 6 | eqtr2d | ⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( 𝑃 ‘ ( ♯ ‘ 𝐹 ) ) = ( lastS ‘ 𝑃 ) ) |