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Proof step 6 from proof square-ls-04 (previous step)

i * i + i < 2, 0 < i ⊦ i * i + 1 < 2

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: int-basic2 name: lst-03 seq: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0 subst: [i, j, i0, j0] → [i * i, i, 2, 1]

Additional information

Used simplifier rules: square-lez-01: ⊦ i * i ≤ 0 ↔ i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i < 1 ↔ i ≤ 0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0); (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j → ¬ j < i; (s) Used terms: i * i, i, 2, 1