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A very neat puzzle! I felt like I needed to deduce and use a range of lemmas and corollaires to solve it.

Before we dive in, a couple of lemmas/corollaries:

Lemma 1: Consider the pattern:

We must have A > D

Corollaries: A cannot be 1 and D cannot be 6

Lemma 2: Consider the pattern:

We must have A+B = D+E

Armed with Lemma 1 we get:

Applying our corollary to column 3 forces

The 2 in R6C3 with some more cells coming as a result of lemma 1

Now

To ensure the horizontal bar is in the correct place in column 2, the 3 must be placed in R5C2

This forces

Row 3 into this unique arrangement with respect to the vertical bar

This is where we bring in lemma 2

R2C3 must be one more than R5C3. The only consecutive pair of numbers that fit are 4 and 5

Let’s go back to lemma 1 to get

Now

The 6 in row 5 can only go in one place to ensure the horizontal bar is in the correct place. This gives a lot more cells via the usual Latin square logic.

Finally, using lemma 2

In row 4 forces the final unique solution (which I slipped up on in my original submission - thanks Dante)

A very neat puzzle! Before we dive in, a couple of lemmas/corollaries:

Lemma 1: Consider the pattern:

We must have A > D

Corollaries: A cannot be 1 and D cannot be 6

Lemma 2: Consider the pattern:

We must have A+B = D+E

Armed with Lemma 1 we get:

Applying our corollary to column 3 forces

The 2 in R6C3 with some more cells coming as a result of lemma 1

Now

To ensure the horizontal bar is in the correct place in column 2, the 3 must be placed in R5C2

This forces

Row 3 into this unique arrangement with respect to the vertical bar

This is where we bring in lemma 2

R2C3 must be one more than R5C3. The only consecutive pair of numbers that fit are 4 and 5

Let’s go back to lemma 1 to get

Now

The 6 in row 5 can only go in one place to ensure the horizontal bar is in the correct place. This gives a lot more cells via the usual Latin square logic.

Finally, using lemma 2

In row 4 forces the final unique solution (which I slipped up on in my original submission - thanks Dante)

A very neat puzzle! I felt like I needed to deduce and use a range of lemmas and corollaires to solve it.

Before we dive in, a couple of lemmas/corollaries:

Lemma 1: Consider the pattern A-B (- is a grey bar, _ is and number). We must have A > B
Corollaries: 1-_ and _-6 are impossible configurations
Lemma 2: Consider the pattern --. The leftmost two numbers must have the same sum as the rightmost 2 numbers

Armed with Lemma 1 we get:

Applying our corollary to column 3 forces

The 2 in R6C3 with some more cells coming as a result of lemma 1

Now

To ensure the horizontal bar is in the correct place in column 2, the 3 must be placed in R5C2

This forces

Row 3 into this unique arrangement with respect to the vertical bar

This is where we bring in lemma 2

R2C3 must be one more than R5C3. The only consecutive pair of numbers that fit are 4 and 5

Let’s go back to lemma 1 to get

Now

The 6 in row 5 can only go in one place to ensure the horizontal bar is in the correct place. This gives a lot more cells via the usual Latin square logic.

Finally, using lemma 2

In row 4 forces the final unique solution (which I slipped up on in my original submission - thanks Dante)

A very neat puzzle! I felt like I needed to deduce and use a range of lemmas and corollaires to solve it.

Before we dive in, a couple of lemmas/corollaries:

Lemma 1: Consider the pattern:

We must have A > D

Corollaries: A cannot be 1 and D cannot be 6

Lemma 2: Consider the pattern:

We must have A+B = D+E

Armed with Lemma 1 we get:

Applying our corollary to column 3 forces

The 2 in R6C3 with some more cells coming as a result of lemma 1

Now

To ensure the horizontal bar is in the correct place in column 2, the 3 must be placed in R5C2

This forces

Row 3 into this unique arrangement with respect to the vertical bar

This is where we bring in lemma 2

R2C3 must be one more than R5C3. The only consecutive pair of numbers that fit are 4 and 5

Let’s go back to lemma 1 to get

Now

The 6 in row 5 can only go in one place to ensure the horizontal bar is in the correct place. This gives a lot more cells via the usual Latin square logic.

Finally, using lemma 2

In row 4 forces the final unique solution (which I slipped up on in my original submission - thanks Dante)

A very neat puzzle! I felt like I needed to deduce and use a range of lemmas and corollaires to solve it.

I’ll try and narrow down the solve path and add it to my solution.

A very neat puzzle! I felt like I needed to deduce and use a range of lemmas and corollaires to solve it.

Before we dive in, a couple of lemmas/corollaries:

Lemma 1: Consider the pattern A-B (- is a grey bar, _ is and number). We must have A > B
Corollaries: 1-_ and _-6 are impossible configurations
Lemma 2: Consider the pattern --. The leftmost two numbers must have the same sum as the rightmost 2 numbers

Armed with Lemma 1 we get:

Applying our corollary to column 3 forces

The 2 in R6C3 with some more cells coming as a result of lemma 1

Now

To ensure the horizontal bar is in the correct place in column 2, the 3 must be placed in R5C2

This forces

Row 3 into this unique arrangement with respect to the vertical bar

This is where we bring in lemma 2

R2C3 must be one more than R5C3. The only consecutive pair of numbers that fit are 4 and 5

Let’s go back to lemma 1 to get

Now

The 6 in row 5 can only go in one place to ensure the horizontal bar is in the correct place. This gives a lot more cells via the usual Latin square logic.

Finally, using lemma 2

In row 4 forces the final unique solution (which I slipped up on in my original submission - thanks Dante)