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Proof step 6 from proof countrefs-lastn (previous step)

¬ # lastn(i→n(countrefs(i1 ' + ints1)), r1 ' + refs2) ≠ # (r0 ' + refs1), # 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 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ n ≤ # refs → # lastn(n, refs) = n
Proof step 5:
¬ i→n(countrefs(i1 ' + ints1)) ≤ # (r1 ' + refs2), ¬ # lastn(i→n(countrefs(i1 ' + ints1)), r1 ' + refs2) ≠ # (r0 ' + refs1), # 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

[speclemmaarg: ,n ≤ # refs → # lastn(n, refs) = n ,[refs, n] → [r1 ' + refs2, i→n(countrefs(i1 ' + ints1))] ,F ,list-del ,references ,lastN-length ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # (refs + refs0) = # refs + # refs0], [CSIMPSEQ : ⊦ # r ' = 0 +1], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPFORWARD : ⊦ r ' + refs = refs0 → r ∈ refs0], [CSIMPSEQ : ⊦ m * (n + 1) = m * n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m + n ≤ m + n0 ↔ n ≤ n0], [CSIMPSEQ : ⊦ m ≤ n → (n ≤ m ↔ m = n)], [CSIMPSEQ : ⊦ m = m0 + 1 + n → m ≠ n])], [EXTRATERMS : (r1 ' + refs2, i→n(countrefs(i1 ' + ints1)))]]