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Proof step 15 from proof sub-less (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 16:
i1 < i + i1, 1 < i, 0 < i1, 0 < j, j < i, 0 < i, i < i + i1 ⊦ j < i + i1, j < i ∧ i < i + i1

Rule argument

pos: left 7

Additional information

Used simplifier rules: gr-def: ⊦ i > j ↔ j < i;% ⊦ i < i + j ↔ 0 < j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i; (s)