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Proof step 6 from proof sub-less (previous step)

i > 0 → i1 < i1 + i, 0 < i, j < i, 0 < j, 1 < i, 0 < i1 ⊦ j < i + i1

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Used simplifier rules: gr-def: ⊦ i > j ↔ j < i;%lst: ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i; from specification int-basic1