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Proof step 1 from proof max_exp-pos-03 (conclusion)

⊦ max_exp(i0, j0) + max_exp(i, j) ≤ 0 ↔ max_exp(i0, j0) = 0 ∧ max_exp(i, j) = 0

User applied rule case distinction

And obtained the following new goals

Proof step 2:
max_exp(i0, j0) + max_exp(i, j) ≤ 0 ⊦ max_exp(i0, j0) = 0 ∧ max_exp(i, j) = 0

Rule argument

pos: right 1

Additional information

Used simplifier rules: max_exp-pos-04: ⊦ ¬ max_exp(i, j) < 0;%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%lel: ⊦ i < i0 ∧ i = i1 → i1 < i0; (forward) from specification oelem[intlista]%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2