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Proof step 5 from proof size-eq (
previous step
)
∀ key. key ∈ ks0 ↔ key ∈ ks1 ⊦ # ks0 = # ks1, ∀ key. key ∈ ks0 ↔ key ∈ ks1
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
∀ key. key ∈ ks0 ↔ key ∈ ks1 ⊦ # ks0 = # ks1, key ∈ ks0 ↔ key ∈ ks1
Rule argument
Additional information