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Proof step 6 from proof test-01 (conclusion)

⊦ ¬ is_fskey(sk) ↔ is_indexkey(sk)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 1:
⊦
Proof step 2:
⊦ is_fskey(fs ')
Proof step 3:
⊦ ¬ is_indexkey(fs ')
Proof step 4:
⊦ ¬ is_fskey(i ')
Proof step 5:
⊦ is_indexkey(i ')

Rule argument

[term-arg: sk]

Additional information

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