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Proof step 5 from proof keys-in (previous step)

∀ sk. sk ∈ fis0 .keys ↔ sk ∈ fis0 ⊦ sk = (sk0 × val) .1 ∨ sk ∈ fis0 .keys ↔ sk ∈ (sk0 × val) ' ∨ sk ∈ fis0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ sk. sk ∈ fis0 .keys ↔ sk ∈ fis0 ⊦ sk ∈ fis0 .keys ↔ sk ∈ fis0, sk = sk0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sk ∈ (sk0 × val) ' ↔ sk = sk0], [CSIMPSEQ : ⊦ (sk × val) .1 = sk])]]