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Proof step 3 from proof countrefs-le-03 (conclusion)

⊦ # refs = i→n(countrefs((i2 + 1) ' + i1 ' + ints1)) ∧ 0 ≤ i2 → ¬ # refs ≤ # refs - i→n(countrefs(i1 ' + ints1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# refs = i→n(countrefs((i2 + 1) ' + i1 ' + ints1)), # refs ≤ # refs - i→n(countrefs(i1 ' + ints1)), 0 ≤ i2 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j])]]