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Proof step 4 from proof max_exp-fac-01 (previous step)

0 < i, 0 ≤ j0, j0 < i, prime(i), ∀ j, i. i→n(j) < i→n(j0) → prime(i) ∧ j < i ∧ 0 ≤ j ∧ 0 < i → max_exp(i, fac(i + j)) = 1 ⊦ max_exp(i, fac(i + j0)) = 1

User applied rule cut formula

And obtained the following new goals

Proof step 5:
0 < i, 0 ≤ j0, j0 < i, prime(i), ∀ j, i. i→n(j) < i→n(j0) → prime(i) ∧ j < i ∧ 0 ≤ j ∧ 0 < i → max_exp(i, fac(i + j)) = 1 ⊦ j0 = 0, max_exp(i, fac(i + j0)) = 1

Rule argument

fma: j0 = 0

Additional information

Used simplifier rules: max_exp-fac-same: ⊦ prime(i) → max_exp(i, fac(i)) = 1;%ax-04: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: j0 = 0