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Proof step 20 from proof add-def-02 (previous step)

gcd(i1, j1) = 1, 0 < j1, 0 < i, 1 < i, 0 < i3, ∀ j2, j1, i4, i3. i→n(i3) + i→n(i4) < i→n(i1) + i→n(i * i0) → 0 < j1 ∧ 0 < j2 → rational(i3, j1) + rational(i4, j2) = rational(i3 * j2 + j1 * i4, j1 * j2) ⊦ rational(i0, i3) + rational(i1, j1) = rational(j1 * i0 + i1 * i3, i3 * j1), i = 0, i3 = 0, j1 = 0

User applied rule apply induction

And obtained the following new goals

Proof step 21:
gcd(i1, j1) = 1, 0 < j1, 0 < i, 1 < i, 0 < i3, ∀ j2, j1, i4, i3. i→n(i3) + i→n(i4) < i→n(i1) + i→n(i * i0) → 0 < j1 ∧ 0 < j2 → rational(i3, j1) + rational(i4, j2) = rational(i3 * j2 + j1 * i4, j1 * j2) ⊦ i→n(i0) + i→n(i1) < i→n(i1) + i→n(i * i0), rational(i0, i3) + rational(i1, j1) = rational(j1 * i0 + i1 * i3, i3 * j1), i = 0, i3 = 0, j1 = 0

Rule argument

subst: [j2, j1, i4, i3] → [j1, i3, i1, i0]

Additional information

Used simplifier rules: add-comm: ⊦ ri + rj = rj + ri;% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i < j → ¬ j < i; Used terms: j1, i3, i1, i0