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Proof step 4 from proof noval-01 (previous step)

is_valuelist(val0), flatten(val1 ++ val0) = noval ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ ∃ val1, val2. flatten(val ++ val0) = val1 ++ val2

Rule argument

[LEMMAARG : exlist, [val, val0] → [val1, val0]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ val ++ val0 ≠ noval])], [EXTRATERMS : (val1, val0)]]