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Proof step 7 from proof countrefs-lastn (previous step)

# 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 1:
⊦ # refs ≠ # refs0 → refs ≠ refs0
Proof step 6:
¬ # 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: ,# refs ≠ # refs0 → refs ≠ refs0 ,[refs0, refs] → [r0 ' + refs1, lastn(i→n(countrefs(i1 ' + ints1)), r1 ' + refs2)] ,F ,list-data ,references ,length-neq ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r0 ' + refs1, lastn(i→n(countrefs(i1 ' + ints1)), r1 ' + refs2))]]