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Proof step 8 from proof ls-even-01 (previous step)

i % 2 = 0, j % 2 = 0, i % 2 = 0 ∧ ~ j % 2 = 0 ∧ 0 ≤ i ∧ 0 ≤ ~ j → (i / 2 < ~ j / 2 ↔ i < ~ j) ⊦ i / 2 < j / 2 ↔ i < j, 0 ≤ i ∧ 0 ≤ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
j % 2 = 0, i % 2 = 0 ⊦ i / 2 < j / 2 ↔ i < j, 0 ≤ i ∧ 0 ≤ j

Rule argument

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%num-ls: ⊦ 0 < j ∧ 0 ≤ i → (i / (* j) < j0 ↔ i < j * j0);%divmult: j ≠ 0 ⊦ i % j = 0 → i / j * j = i;% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ ~ i * j = ~(i * j); (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)% ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)% ⊦ ~ i = 0 ↔ i = 0; (s)