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Proof step 8 from proof countrefs-lastn (conclusion)

⊦ # refs2 = i→n(countrefs(i1 ' + ints1)) * i→n(i2) + i→n(countrefs(i1 ' + ints1)) ∧ lastn(i→n(countrefs(i1 ' + ints1)), r1 ' + refs2) = r0 ' + refs1 ∧ 0 ≤ i2 → ¬ # refs2 ≤ # refs1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# refs2 = i→n(countrefs(i1 ' + ints1)) * i→n(i2) + i→n(countrefs(i1 ' + ints1)), lastn(i→n(countrefs(i1 ' + ints1)), r1 ' + refs2) = r0 ' + refs1, # refs2 ≤ # refs1, 0 ≤ i2 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)], [CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPFORWARD : ⊦ r ' + refs = refs0 → r ∈ refs0])]]