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Proof step 77 from proof square-neq-any-pos-02 (previous step)

∀ i0, j, j0. i→n(i0) < i→n(1) → 0 < i0 ∧ 0 < j ∧ 0 < j0 ∧ gcd(j, j0) = 1 ∧ j * j = i0 * j0 * j0 → (∃ i1. i0 = i1 * i1) ⊦ ∃ i0. 1 = i0 * i0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: 1 abort: F discard: T computed: F

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 1