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

⊦ fs[ax] = nd ' ↔ (∃ adr. ax = adr ' ∧ fs[adr] = nd)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
fs[ax] = nd ' ⊦ ∃ adr. ax = adr ' ∧ fs[adr] = nd

Rule argument

Additional information

Used simplifier rules: read-one: ⊦ fs[adr '] = fs[adr] ';% ⊦ nd ' = nd0 ' ↔ nd = nd0; (s)