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Proof step 6 from proof put-rkvs-append (previous step)

∀ st, rkvs0. st[rkvs1 + rkvs0] = (st[rkvs1][rkvs0]) ⊦ st[(rkv + rkvs1) + rkvs0] = (st[rkv + rkvs1][rkvs0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ st[rkv ' + rkvs] = (st[rkv .1, rkv .2][rkvs])
Proof step 5:
∀ st, rkvs0. st[rkvs1 + rkvs0] = (st[rkvs1][rkvs0]) ⊦ (st[rkv .1, rkv .2][rkvs1 + rkvs0]) = (st[rkv .1, rkv .2][rkvs1][rkvs0])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (rkvs + rkvs0) + rkvs1 = rkvs + rkvs0 + rkvs1], [CSIMPSEQ : ⊦ rkv + rkvs = rkv ' + rkvs])]]