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Proof step 4 from proof lookup-put-rkvs-diff-03 (previous step)

str ≠ str0 ⊦ (st[((r -- mkfs(class, ty, str .field) ') × val) ' + rkvs][r0 -- mkfs(class0, ty0, str0 .field) ']) = (st[rkvs][r0 -- mkfs(class0, ty0, str0 .field) '])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ st[(rk × val) ' + rkvs] = (st[rk, val][rkvs])
Proof step 3:
⊦ str = str0, (st[r -- mkfs(class, ty, str .field) ', val][rkvs][r0 -- mkfs(class0, ty0, str0 .field) ']) = (st[rkvs][r0 -- mkfs(class0, ty0, str0 .field) '])

Rule argument

[emptyarg]

Additional information

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