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Proof step 10 from proof in-list-put (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 9:
∀ r, st, r0, sk, val. (r -- sks0 ∈ st[r0 -- sk, val] ↔ r -- sks0 ∈ st) ∨ sk ∈ sks0 ⊦ sks0 = [], r -- (sk0 ' + sks0) ∈ st[r0 -- sk, val] ↔ r -- (sk0 ' + sks0) ∈ st, sk = sk0, sks0 = [], sk ∈ sks0

Rule argument

[fma-arg: sks0 = []]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (sks0 = [])]]