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Proof step 18 from proof countrefs-le (previous step)

# refs = n + i→n(countrefs(ints)) ∧ i→n(countrefs(ints)) ≤ # refs, # refs = i→n(countrefs(i ' + ints)), # refs ≤ n, 0 < i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# refs = n + i→n(countrefs(ints)), # refs = i→n(countrefs(i ' + ints)), i→n(countrefs(ints)) ≤ # refs, # refs ≤ n, 0 < i, i→n(countrefs(ints)) ≤ n ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1])]]