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Proof step 4 from proof union-at-left (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ st ∪ st0[rk, val] = (st ∪ st0)[rk, val]
Proof step 3:
∀ st, rk. (st ∪ st1)[rk] = st[rk] ∨ rk ∈ st1 ⊦ rk = rk0, (st ∪ st1)[rk] = st[rk], rk ∈ st1

Rule argument

[emptyarg]

Additional information

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