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Proof step 3 from proof countrefs-rec-01 (conclusion)

⊦ 0 ≤ i → countrefs((i + 1) ' + i0 ' + ints) = 1 + (i + 1) * countrefs(i0 ' + ints)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i ⊦ countrefs((i + 1) ' + i0 ' + ints) = 1 + (i + 1) * countrefs(i0 ' + ints)

Rule argument

[emptyarg]

Additional information

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