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Proof step 29 from proof ls-even-01 (previous step)

i % 2 = 0, j % 2 = 0, 0 < j / 2 + i / 2 ↔ 0 < j + i, i < 0, i < j, 0 < j, j / 2 ≤ i / 2 ⊦

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: ls-even seq: ⊦ i % 2 = 0 ∧ j % 2 = 0 ∧ 0 ≤ i ∧ 0 ≤ j → (i / 2 < j / 2 ↔ i < j) subst: [i, j] → [0, j]

Additional information

Used simplifier rules: lstwozero: ⊦ i < 0 → ¬ 0 < i / 2;%div-ls-zero: 0 < j ⊦ i / (* j) < 0 ↔ i + j ≤ 0;% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs% ⊦ i + j < j ↔ i < 0; (s)% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i + j ≤ i ↔ j ≤ 0; (s)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s) Used terms: 0, j