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Proof step 6 from proof est-4-big-01 (previous step)

(2 * i) / 3 < 2 * i % j, 0 < i, 2 * i < 3 * j, j ≤ i, 2 < i, 2 * i < j * j, j ≤ 2 * i % j, 0 ≤ i, j = 2 ∨ j = 3 ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 9

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%nummult-ls: ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i); from specification int-div%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%num-between: i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; from specification int-basic2