Metamath Proof Explorer


Theorem problem3

Description: Practice problem 3. Clues: eqcomi eqtri subaddrii recni 4re 3re 1re df-4 addcomi . (Contributed by Filip Cernatescu, 16-Mar-2019) (Proof modification is discouraged.)

Ref Expression
Hypotheses problem3.1 A
problem3.2 A + 3 = 4
Assertion problem3 A = 1

Proof

Step Hyp Ref Expression
1 problem3.1 A
2 problem3.2 A + 3 = 4
3 4re 4
4 3 recni 4
5 3re 3
6 5 recni 3
7 1re 1
8 7 recni 1
9 df-4 4 = 3 + 1
10 9 eqcomi 3 + 1 = 4
11 4 6 8 10 subaddrii 4 3 = 1
12 11 eqcomi 1 = 4 3
13 6 1 addcomi 3 + A = A + 3
14 13 2 eqtri 3 + A = 4
15 4 6 1 14 subaddrii 4 3 = A
16 12 15 eqtri 1 = A
17 16 eqcomi A = 1