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Proof step 1 from proof keys-eq-one (conclusion)

⊦ ndx.keys = key ' ↔ (∃ nd. ndx = nd ' ∧ nd.key = key)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
ndx.keys = key ' ⊦ ∃ nd. ndx = nd ' ∧ nd.key = key

Rule argument

Additional information

Used simplifier rules: keys-one: ⊦ nd '.keys = nd.key ';% ⊦ key ' = key0 ' ↔ key = key0; (s)