Proof: Rearrange Sum 6

Let's prove the following theorem:

((((a + b) + c) + d) + e) + f = ((((a + e) + b) + d) + f) + c

Proof:

View as a tree | View dependent proofs | Try proving it

Proof Table
# Claim Reason
1 (((a + b) + c) + d) + e = (((a + b) + c) + e) + d (((a + b) + c) + d) + e = (((a + b) + c) + e) + d
2 ((a + b) + c) + e = ((a + b) + e) + c ((a + b) + c) + e = ((a + b) + e) + c
3 (a + b) + e = (a + e) + b (a + b) + e = (a + e) + b
4 ((a + b) + e) + c = ((a + e) + b) + c if (a + b) + e = (a + e) + b, then ((a + b) + e) + c = ((a + e) + b) + c
5 ((a + b) + c) + e = ((a + e) + b) + c if ((a + b) + e) + c = ((a + e) + b) + c and ((a + b) + c) + e = ((a + b) + e) + c, then ((a + b) + c) + e = ((a + e) + b) + c
6 (((a + b) + c) + e) + d = (((a + e) + b) + c) + d if ((a + b) + c) + e = ((a + e) + b) + c, then (((a + b) + c) + e) + d = (((a + e) + b) + c) + d
7 (((a + b) + c) + d) + e = (((a + e) + b) + c) + d if (((a + b) + c) + e) + d = (((a + e) + b) + c) + d and (((a + b) + c) + d) + e = (((a + b) + c) + e) + d, then (((a + b) + c) + d) + e = (((a + e) + b) + c) + d
8 (((a + e) + b) + c) + d = (((a + e) + b) + d) + c (((a + e) + b) + c) + d = (((a + e) + b) + d) + c
9 (((a + b) + c) + d) + e = (((a + e) + b) + d) + c if (((a + e) + b) + c) + d = (((a + e) + b) + d) + c and (((a + b) + c) + d) + e = (((a + e) + b) + c) + d, then (((a + b) + c) + d) + e = (((a + e) + b) + d) + c
10 ((((a + b) + c) + d) + e) + f = ((((a + e) + b) + d) + c) + f if (((a + b) + c) + d) + e = (((a + e) + b) + d) + c, then ((((a + b) + c) + d) + e) + f = ((((a + e) + b) + d) + c) + f
11 ((((a + e) + b) + d) + c) + f = ((((a + e) + b) + d) + f) + c ((((a + e) + b) + d) + c) + f = ((((a + e) + b) + d) + f) + c
12 ((((a + b) + c) + d) + e) + f = ((((a + e) + b) + d) + f) + c if ((((a + e) + b) + d) + c) + f = ((((a + e) + b) + d) + f) + c and ((((a + b) + c) + d) + e) + f = ((((a + e) + b) + d) + c) + f, then ((((a + b) + c) + d) + e) + f = ((((a + e) + b) + d) + f) + c

Comments

Please log in to add comments