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

j0 < 0 ∨ 0 < j0 ⊦ j0 = ~1, 0 = j0, j0 +1 < 0, 0 < j0 +1

User applied rule case distinction

And obtained the following new goals

Proof step 6:
j0 < 0 ⊦ j0 = ~1, 0 = j0, j0 +1 < 0, 0 < j0 +1
Proof step 8:
0 < j0 ⊦ j0 = ~1, 0 = j0, j0 +1 < 0, 0 < j0 +1

Rule argument

pos: left 1

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