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Proof step 11 from proof sub-less (
previous step
)
j < i ∧ i < i + i1 → j < i + i1, i1 < i + i1, 0 < i, j < i, 0 < j, 1 < i, 0 < i1 ⊦ j < i + i1
User applied rule case distinction
And obtained the following new goals
Proof step 12
:
i1 < i + i1, 0 < i, j < i, 0 < j, 1 < i, 0 < i1 ⊦ j < i ∧ i < i + i1, j < i + i1
Rule argument
pos: left 1
Additional information