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Proof step 11 from proof noval-01 (previous step)

is_valuelist(val), flatten(val) = noval ⊦ val = noval

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ is_valuelist(val) ↔ (∃ val0, val1. val = val0 ++ val1)
Proof step 10:
∃ val0, val1. val = val0 ++ val1, flatten(val) = noval ⊦ val = noval

Rule argument

[speclemmaarg: ,is_valuelist(val) ↔ (∃ val0, val1. val = val0 ++ val1) ,[val] → [val] ,F ,javavalue , ,ex-02 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val)]]