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Proof step 6 from proof max_exp-fac-same (previous step)

0 < i ∧ 0 < fac(i - 1) ∧ prime(i), 0 < i, prime(i) ⊦ max_exp(i, i) + max_exp(i, fac(i - 1)) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: max_exp-same: ⊦ prime(i) → max_exp(i, i) = 1;%max_exp-zero-02: ⊦ prime(i) ∧ 0 < i0 → (max_exp(i, i0) = 0 ↔ ¬ i | i0);%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) 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%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs