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Proof step 14 from proof no-oc (conclusion)
| ⊦ #oc(n, A, m0) = 0 ↔ (∀ m. m < m0 → A[m] ≠ n) |
User applied rule structural induction
And obtained the following new goals
- Proof step 13:
∀ n, A. (#oc(n, A, m) = 0 ↔ (∀ m1. m1 < m → A[m1] ≠ n))
⊦ #oc(n, A, m +1) = 0 ↔ (∀ m1. m1 < m +1 → A[m1] ≠ n)
Rule argument
indvar:m0Additional information
[Concretesimprules((Csimpgen(nat freely generated by 0, +1;)
Csimpnamedseq(array-perm,array_of_nat,numoc-base,
⊦ #oc(n, A, 0) = 0)
Csimpnamedseq(nat-basic1,,zero-01,
⊦ ¬ m < 0)))]