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Proof step 9 from proof at-put-last-other (previous step)

sks0 + 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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ 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

[emptyarg]

Additional information

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