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Proof step 11 from proof lookup-put-rkvs (previous step)

∀ rk, rk0, st, val. rk = rk0 ∨ (st[rk, val][rkvs0][rk0]) = (st[rkvs0][rk0]) ⊦ rk = rk0, (st[rk, val][rkv + rkvs0][rk0]) = (st[rkv + rkvs0][rk0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ st[rkvs + rkvs0] = (st[rkvs][rkvs0])
Proof step 4:
⊦ st[rkv '] = st[rkv .1, rkv .2]
Proof step 10:
∀ rk, rk0, st, val. rk = rk0 ∨ (st[rk, val][rkvs0][rk0]) = (st[rkvs0][rk0]) ⊦ rk = rk0, (st[rk, val][rkv .1, rkv .2][rkvs0][rk0]) = (st[rkv .1, rkv .2][rkvs0][rk0])

Rule argument

[emptyarg]

Additional information

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