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Proof step 3 from proof test-02 (previous step)

is_indexkey(sk), is_fskey(sk) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ is_indexkey(sk) ↔ (∃ i. sk = i ')
Proof step 2:
⊦ ¬ is_fskey(i ')

Rule argument

[crewrite-arg: ex-04, F, [sk] → [sk]]

Additional information

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