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Proof step 8 from proof at-put-last-other (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 7:
r0 = st[st[r -- sk] .refval -- sks0] .refval ∨ sks0 = [] ∨ (st[st[r -- sk] .refval -- (sks0 + sk0 '), val][r0 -- sk0]) = st[r0 -- sk0], ∀ r0, st, r, sk0, val. r0 = st[r -- sks0] .refval ∨ sks0 = [] ∨ (st[r -- (sks0 + sk0 '), val][r0 -- sk0]) = st[r0 -- sk0] ⊦ r0 = st[r -- (sk ' + sks0)] .refval, (st[st[r -- sk] .refval -- (sks0 + sk0 '), val][r0 -- sk0]) = st[r0 -- sk0]

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r0, st, st[r -- sk] .refval, sk0, val)]]