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

∀ r, r0, st, sk, val. r = r0 ∨ (hasKeys(r, sks0, st[r0 -- sk, val]) ↔ hasKeys(r, sks0, st)) ⊦ r = r0, hasKeys(r, sk0 + sks0, st[r0 -- sk, val]) ↔ hasKeys(r, sk0 + 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 6:
∀ r, r0, st, sk, val. r = r0 ∨ (hasKeys(r, sks0, st[r0 -- sk, val]) ↔ hasKeys(r, sks0, st)) ⊦ r = r0, r -- sk0 ∈ st ∧ hasKeys(r, sks0, st[r0 -- sk, val]) ↔ hasKeys(r, sk0 ' + sks0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ r ≠ r0 → ( r -- sk ∈ st[r0 -- sk0, val] ↔ r -- sk ∈ st)], [CSIMPSEQ : ⊦ sk + sks = sk ' + sks])]]