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

∀ 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

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (st[r -- sk0] .refval, st, r0, sk, val) abort:F , discard:F , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)])], [EXTRATERMS : (st[r -- sk0] .refval, st, r0, sk, val)]]