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Proof step 6 from proof update-keyin-01 (previous step)

r = r0 ∧ sk = sk0 ∨ r -- sk ∈ update(r0, fis0, st), r -- sk ∈ update(r0, fis0, st) ↔ r -- sk ∈ st ∨ r = r0 ∧ sk ∈ fis0, ∀ r, sk, r0, st. r -- sk ∈ update(r0, fis0, st) ↔ r -- sk ∈ st ∨ r = r0 ∧ sk ∈ fis0 ⊦ r -- sk ∈ st ∨ r = r0 ∧ (sk = sk0 ∨ sk ∈ fis0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
r -- sk ∈ update(r0, fis0, st) ↔ r = r0 ∧ sk ∈ fis0, r = r0 ∧ sk = sk0 ∨ r = r0 ∧ sk ∈ fis0, ∀ r, sk, r0, st. r -- sk ∈ update(r0, fis0, st) ↔ r -- sk ∈ st ∨ r = r0 ∧ sk ∈ fis0 ⊦ r -- sk ∈ st, r = r0 ∧ (sk = sk0 ∨ sk ∈ fis0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]