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Proof step 7 from proof size-eq (previous step)

key ∈ ks0, ∀ key. key ∈ ks0 ↔ key ∈ ks1 ⊦ # ks0 = # ks1, key ∈ ks1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
key ∈ ks0, ∀ key. key ∈ ks0 ↔ key ∈ ks1 ⊦ # ks0 = # ks1, # ks0 = 0, ks0 = ∅, key ∈ ks1

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ ks ≠ ∅ → # ks ≠ 0; (forward) from specification set-basic[node-index]%ne-01: ⊦ key ∈ ks → ks ≠ ∅; (forward) from specification set-basic[node-index]