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Proof step 15 from proof lst-02 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 16:
0 < i +1 ⊦ i = ~1, i = 0, i < 0, 0 < i
Proof step 21:
i +1 < 0 ⊦ i = ~1, i = 0, i < 0, 0 < i

Rule argument

pos: left 1

Additional information