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Proof step 10 from proof countrefs-rec-02 (previous step)

# (i2 ' + i1 ' + ints1) > 1 ∧ (i2 ' + i1 ' + ints1) .first > 0, 0 ≤ i2 ⊦ i→n(1 + (i2 ' + i1 ' + ints1) .first * countrefs((i2 ' + i1 ' + ints1) .rest)) = i→n(countrefs(i1 ' + ints1)) * i→n(i2) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPSEQ : ⊦ i > j ↔ j < i], [CSIMPSEQ : ⊦ # ints = 0 ↔ ints = []], [CSIMPSEQ : ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : n < n0 ⊦ ((* n)) ≠ m0 + ((* n0))], [CSIMPSEQ : ⊦ 1 < m ↔ ¬ (m = 0 ∨ m = 1)], [CSIMPSEQ : ⊦ m > n ↔ n < m], [CSIMPSEQ : ⊦ # i ' = 0 +1], [CSIMPSEQ : ⊦ # (ints + ints0) = # ints + # ints0], [CSIMPSEQ : ⊦ (i ' + ints) .first = i], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ (i ' + ints) .rest = ints])]]