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Proof step 5 from proof in-list-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ r, st, r0, sk, val. (r -- sks0 ∈ st[r0 -- sk, val] ↔ r -- sks0 ∈ st) ∨ sk ∈ sks0 ⊦ sks0 = [], r -- sk0 ∈ st[r0 -- sk, val] ∧ st[r -- sk0] .refval -- sks0 ∈ st[r0 -- sk, val] ↔ r -- sk0 ∈ st ∧ st[r -- sk0] .refval -- sks0 ∈ st, sk = sk0, sk ∈ sks0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]