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Proof step 10 from proof hasKeys-put-01 (previous step)

∀ rk, st, r, val. rk ∈ st → (hasKeys(r, sks0, st[rk, val]) ↔ hasKeys(r, sks0, st)), rk ∈ st ⊦ hasKeys(r, sk + sks0, st[rk, val]) ↔ hasKeys(r, sk + sks0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ hasKeys(r, sks + sks0, st) ↔ hasKeys(r, sks, st) ∧ hasKeys(r, sks0, st)
Proof step 3:
⊦ hasKeys(r, sk ', st) ↔ r -- sk ∈ st
Proof step 9:
rk ∈ st, ∀ rk, st, r, val. rk ∈ st → (hasKeys(r, sks0, st[rk, val]) ↔ hasKeys(r, sks0, st)) ⊦ r -- sk ∈ st[rk, val] ∧ hasKeys(r, sks0, st[rk, val]) ↔ hasKeys(r, sk ' + sks0, st)

Rule argument

[emptyarg]

Additional information

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