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Proof step 15 from proof countrefs-le (previous step)

# refs = i→n(countrefs(i ' + ints)), # refs ≤ # refs - i→n(countrefs(ints)), 0 < i ⊦ i→n(countrefs(ints)) ≤ # refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ countrefs(ints) ≠ 0
Proof step 14:
# refs = i→n(countrefs(i ' + ints)), # refs < i→n(countrefs(ints)), 0 < i, # refs ≤ # refs - i→n(countrefs(ints)), abs(countrefs(i ' + ints)) < abs(countrefs(ints)) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j)], [CSIMPSEQ : ⊦ i→n(i) = 0 ↔ i = 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ m < n ∧ m = n0 → n0 < n], [CSIMPSEQ : ⊦ ¬ m ≤ n ↔ n < m])]]