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Proof step 8 from proof ri-cons-truncnode (previous step)

fs[ri[key0]] = truncnode(key0, m), key0 ∈ ri, ri[key0] ∈ fs, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ key0.data?, key0.dentry?, key0.inode?

User applied rule constructor cut

And closed the goal!

Rule argument

var: key0

Additional information

Used simplifier rules: key freely generated by inodekey, datakey, dentrykey;;% ⊦ ¬ dentrykey(m, str).inode?; (s)% ⊦ dentrykey(m, str).dentry?; (s)% ⊦ ¬ datakey(m, m0).inode?; (s)% ⊦ ¬ datakey(m, m0).dentry?; (s)% ⊦ datakey(m, m0).data?; (s)% ⊦ inodekey(m).inode?; (s)