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Proof step 10 from proof getcountdims (previous step)

∀ i. n→i(# getdims(ints0)) + countdims(ints0, i) = n→i(# ints0) + i ⊦ n→i(# getdims(i0 + ints0)) + countdims(i0 + ints0, i) = n→i(# (i0 + ints0)) + i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ i. n→i(# getdims(ints0)) + countdims(ints0, i) = n→i(# ints0) + i ⊦ countdims(i0 ' + ints0, i) + n→i(# getdims(i0 ' + ints0)) = (1 + n→i(# ints0)) + i

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ n→i(m + n) = n→i(m) + n→i(n)], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ # i ' = 0 +1], [CSIMPSEQ : ⊦ # (ints + ints0) = # ints + # ints0], [CSIMPSEQ : ⊦ i + ints = i ' + ints])]]