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Proof step 5 from proof two-02 (previous step)

3 = 2 + 1, 3 + m0 + m0 + m0 = 2 ⊦ m0 + m0 + m0 = 2

User applied rule insert equation

And obtained the following new goals

Proof step 6:
(2 + 1) + m0 + m0 + m0 = 2 ⊦ m0 + m0 + m0 = 2

Rule argument

pos: left 1 flags: discard

Additional information