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Proof step 7 from proof union-at-right-01 (previous step)

∀ rk, st. rk ∈ st1 → (st ∪ st1)[rk] = st1[rk], rk ∈ st1[rk0, val] ⊦ (st ∪ st1[rk0, val])[rk] = (st1[rk0, val][rk])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ st ∪ st0[rk, val] = (st ∪ st0)[rk, val]
Proof step 6:
rk ∈ st1[rk0, val], ∀ rk, st. rk ∈ st1 → (st ∪ st1)[rk] = st1[rk] ⊦ ((rk = rk0 ⊃ val; (st ∪ st1)[rk])) = ((rk = rk0 ⊃ val; st1[rk]))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))])]]