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Proof step 13 from proof del-this (
previous step
)
⊦ key0 ∈ ri[key, adr] ⊖ key ↔ key0 ∈ ri ⊖ key, (key0 ∈ ri[key, adr] ⊖ key ↔ key0 ∈ ri ⊖ key) ∧ (key0 ∈ ri[key, adr] ⊖ key → (ri[key, adr] ⊖ key)[key0] = (ri ⊖ key)[key0])
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 14
:
⊦ key0 ∈ ri[key, adr] ⊖ key ↔ key0 ∈ ri ⊖ key
Rule argument
Additional information