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Proof step 6 from proof del-put-again (previous step)

inodekey(m) ∈ ks ⊦ # ks = 0, ks = ∅, (key ∈ ri[inodekey(m), adr] ⊖ ks ↔ key ∈ ri ⊖ ks) ∧ (key ∈ ri[inodekey(m), adr] ⊖ ks → (ri[inodekey(m), adr] ⊖ ks)[key] = (ri ⊖ ks)[key])

User applied rule cut formula

And obtained the following new goals

Proof step 7:
key ∈ ri[inodekey(m), adr] ⊖ ks ↔ key ∈ ri ⊖ ks, inodekey(m) ∈ ks ⊦ # ks = 0, ks = ∅, (key ∈ ri[inodekey(m), adr] ⊖ ks ↔ key ∈ ri ⊖ ks) ∧ (key ∈ ri[inodekey(m), adr] ⊖ ks → (ri[inodekey(m), adr] ⊖ ks)[key] = (ri ⊖ ks)[key])
Proof step 18:
inodekey(m) ∈ ks ⊦ # ks = 0, key ∈ ri[inodekey(m), adr] ⊖ ks ↔ key ∈ ri ⊖ ks, ks = ∅, (key ∈ ri[inodekey(m), adr] ⊖ ks ↔ key ∈ ri ⊖ ks) ∧ (key ∈ ri[inodekey(m), adr] ⊖ ks → (ri[inodekey(m), adr] ⊖ ks)[key] = (ri ⊖ ks)[key])

Rule argument

fma: key ∈ ri[inodekey(m), adr] ⊖ ks ↔ key ∈ ri ⊖ ks

Additional information

Used formulas: key ∈ ri[inodekey(m), adr] ⊖ ks ↔ key ∈ ri ⊖ ks