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Proof step 3 from proof ex-modeq-half-all (previous step)

∃ i1. ≡mod(i1, i, j0) ∧ abs(i1) ≤ j0 / 2, 0 < j0, odd(j0), i2 ^2 + i1 ^2 + i0 ^2 + i ^2 = j0 * j ⊦ ∃ i3, i4, i5, i6. ≡mod(i3, i, j0) ∧ abs(i3) ≤ j0 / 2 ∧ ≡mod(i4, i0, j0) ∧ abs(i4) ≤ j0 / 2 ∧ ≡mod(i5, i1, j0) ∧ abs(i5) ≤ j0 / 2 ∧ ≡mod(i6, i2, j0) ∧ abs(i6) ≤ j0 / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i2 ^2 + i1 ^2 + i0 ^2 + i ^2 = j0 * j, ≡mod(i3, i, j0), abs(i3) ≤ j0 / 2, odd(j0), 0 < j0 ⊦ ∃ i3, i4, i5, i6. ≡mod(i3, i, j0) ∧ abs(i3) ≤ j0 / 2 ∧ ≡mod(i4, i0, j0) ∧ abs(i4) ≤ j0 / 2 ∧ ≡mod(i5, i1, j0) ∧ abs(i5) ≤ j0 / 2 ∧ ≡mod(i6, i2, j0) ∧ abs(i6) ≤ j0 / 2

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs