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Proof step 3 from proof countmore2 (conclusion)

0 < i ⊦ countrefs2(i ' + i0 ' + ints) = i + i * countrefs2(i0 ' + ints)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < i ⊦ countrefs2(i ' + i0 ' + ints) = i * (countrefs2(i0 ' + ints) + 1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i * j = j * i], [CSIMPSEQ : ⊦ i * j * j0 = (i * j) * j0], [CSIMPSEQ : ⊦ i * j + i = i * (j + 1)])]]