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Proof step 4 from proof in-put-keys (previous step)

sks0 + sk ' ≠ [], ∀ sk, sk0, r0, st, r, val. sk = sk0 ∨ (r0 -- sk0 ∈ st[r -- (sks0 + sk '), val] ↔ r0 -- sk0 ∈ st) ⊦ sk = sk0, r0 -- sk0 ∈ st[st[r -- sk1] .refval -- (sks0 + sk '), val] ↔ r0 -- sk0 ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ sk, sk0, r0, st, r, val. sk = sk0 ∨ (r0 -- sk0 ∈ st[r -- (sks0 + sk '), val] ↔ r0 -- sk0 ∈ st) ⊦ sk = sk0, r0 -- sk0 ∈ st[st[r -- sk1] .refval -- (sks0 + sk '), val] ↔ r0 -- sk0 ∈ st

Rule argument

[emptyarg]

Additional information

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