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Proof step 4 from proof four-square-congmod (previous step)

i2 ^2 + i1 ^2 + i0 ^2 + i ^2 = j0 * j, ≡mod(i3, i2, j0), ≡mod(i4, i1, j0), ≡mod(i5, i0, j0), ≡mod(i6, i, j0), 1 < j, 0 < j, 0 < j0, 0 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ⊦ j0 = 1, j = 2, j0 | abs(i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 - 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i2 ^2 + i1 ^2 + i0 ^2 + i ^2 = j0 * j, ≡mod(i3, i2, j0), ≡mod(i4, i1, j0), ≡mod(i5, i0, j0), ≡mod(i6, i, j0), 1 < j, 0 < j, 0 < j0, 0 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ⊦ j0 = 1, j = 2, j0 | i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2

Rule argument

Additional information

Used simplifier rules: abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%sub-zero-01: ⊦ i - 0 = i; from specification int-abs