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Proof step 12 from proof lagrange-lemma-smallest (previous step)

i1 ^2 + i2 ^2 + i3 ^2 + i4 ^2 = j1 * j, i ^2 + i0 ^2 + i5 ^2 + i6 ^2 = j0 * j, 1 < j, prime(j), 0 < j, j1 < j, 0 < j0, j0 < j1, 1 < j1, 0 < j1, j0 < j, ∀ j0, i1, i2, i3, i4, j. i→n(j0) < i→n(j1) → i1 ^2 + i2 ^2 + i3 ^2 + i4 ^2 = j0 * j ∧ j0 < j ∧ 0 < j0 ∧ 0 < j ∧ prime(j) ∧ 1 < j → j = 2 ∨ (∃ i, i0, i1, i2, j0. i ^2 + i0 ^2 + i1 ^2 + i2 ^2 = j0 * j ∧ 0 < j0 ∧ j0 < j ∧ (∀ i3, i4, i5, i6, j1. 0 < j1 ∧ j1 < j0 → i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≠ j1 * j)) ⊦ j = 2, ∃ i, i0, i1, i2, j0. i ^2 + i0 ^2 + i1 ^2 + i2 ^2 = j0 * j ∧ 0 < j0 ∧ j0 < j ∧ (∀ i3, i4, i5, i6, j1. 0 < j1 ∧ j1 < j0 → i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≠ j1 * j)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [j0, i1, i2, i3, i4, j] → [j0, i, i0, i5, i6, j]

Additional information

Used simplifier rules: labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs: ⊦ 1 < i → abs(i) = i; from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: j0, i, i0, i5, i6, j