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Proof step 3 from proof zero-ge-add (previous step)

j1 < j, 0 < j, 0 ≤ j1 ⊦ 0 ≤ i0 * j + j1 ↔ 0 ≤ i0

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ i ≤ j ↔ ¬ j < i; (ss)%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i + j < j ↔ i < 0; (s)% ⊦ i ≤ j → (j < i ↔ false); (as)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2