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Proof step 3 from proof sub-less (previous step)

⊦ 0 < i ∧ 0 < j ∧ 0 < i1 → i + i1 ≤ j → i ≤ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 < i, j < i, 0 < j, 1 < i, 0 < i1 ⊦ 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;%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1