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Proof step 7 from proof hasKeys-app (previous step)

∀ r, sks0, st. hasKeys(r, sks1 + sks0, st) ↔ hasKeys(r, sks1, st) ∧ hasKeys(r, sks0, st) ⊦ hasKeys(r, (sk + sks1) + sks0, st) ↔ hasKeys(r, sk + sks1, st) ∧ hasKeys(r, sks0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ hasKeys(r, sk ' + sks, st) ↔ r -- sk ∈ st ∧ hasKeys(r, sks, st)
Proof step 6:
∀ r, sks0, st. hasKeys(r, sks1 + sks0, st) ↔ hasKeys(r, sks1, st) ∧ hasKeys(r, sks0, st) ⊦ r -- sk ∈ st ∧ hasKeys(r, sks1 + sks0, st) ↔ hasKeys(r, sk ' + sks1, st) ∧ hasKeys(r, sks0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sk + sks = sk ' + sks])]]