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Proof step 7 from proof sub-less (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 8:
0 < i, j < i, 0 < j, 0 < i1, 1 < i, i1 < i1 + i ⊦ j < i + i1

Rule argument

pos: left 6

Additional information

Used simplifier rules: gr-def: ⊦ i > j ↔ j < i;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i; (s)