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Proof step 3 from proof update-in-01 (conclusion)

⊦ r ≠ r0 → (r ∈ update(r0, fis, st) ↔ r ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ r ∈ update(r0, fis, st) ↔ r ∈ st ∨ r = r0 ∧ fis ≠ []
Proof step 2:
⊦ r ∈ st → r ∈ update(r0, fis, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]