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Proof step 2 from proof boolval-xtra-01 (previous step)

val .boolval ⊦ is_boolvalue(val)

User applied rule insert lemma

And obtained the following new goals

Proof step 1:
⊦ ¬ is_boolvalue(val) → (val .boolval ↔ false)

Rule argument

[LEMMAARG : boolval-xtra, [val] → [val]]

Additional information

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