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Henry Bushby Transcription pages

VM533B94v4-p252.jpg

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Iz 8b/8c even. last=1[sideways U]b+1=1[sideways U]9.
[diagram] c=8 unaltered
z=3 = z+1
1[sideways U]b-4 = 1[sideways U]4
[horizontal line]
[diagram] 1[sideways U]b-3 = 1[sideways U]5
c=4 = 8-4 [bracket spanning lines above and below] gives Iz 4b/4c
z=2, unaltered
[horizontal line]
[diagram] 1[sideways U]b-2=1[sideways U]6
c=8 unaltered
z=1.
[angled pencil note: composite knot
5+3]
[horizontal line]
[diagram] 1[sideways U]b-1 = 1[sideways U]7
c=5 = 8-3 5c T.H.
z=1

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