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Proof step 19 from proof countrefs-le (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
# refs = i→n(countrefs(i ' + ints)), # refs ≤ # refs - i→n(countrefs(ints)), 0 < i ⊦ i→n(countrefs(ints)) ≤ # refs
Proof step 18:
# refs = n + i→n(countrefs(ints)) ∧ i→n(countrefs(ints)) ≤ # refs, # refs = i→n(countrefs(i ' + ints)), # refs ≤ n, 0 < i ⊦

Rule argument

[speclemmaarg: n ≤ m ,n0 = m - n ↔ m = n0 + n ,[m, n, n0] → [# refs, i→n(countrefs(ints)), n] ,T ,nat , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≤ m ⊦ n0 = m - n ↔ m = n0 + n])], [EXTRATERMS : (# refs, i→n(countrefs(ints)), n)]]