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Proof step 4 from proof sub-less (previous step)

0 < i1, 0 < j, 0 < i, i + i1 ≤ j ⊦ i ≤ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < i, j < i, 0 < j, 0 < i1, 1 < i ⊦ j < i + i1

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%numlsf-01: ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j;% ⊦ i < j → ¬ j < i; (s)%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1