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Proof step 7 from proof union-in (previous step)

rk × (st ∪ st0)[rk] ∈ st ∪ st0 ↔ rk × (st ∪ st0)[rk] ∈ st0 ∨ ¬ rk ∈ st0 ∧ rk × (st ∪ st0)[rk] ∈ st ⊦ rk ∈ st ∪ st0 ↔ rk ∈ st ∨ rk ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ rk ∈ st0 → (st ∪ st0)[rk] = st0[rk]
Proof step 6:
rk ∈ st ∪ st0 ↔ rk ∈ st0 ∨ rk ∈ st ∧ st[rk] = (st ∪ st0)[rk] ⊦ rk ∈ st ∪ st0 ↔ rk ∈ st ∨ rk ∈ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk × val ∈ st ↔ rk ∈ st ∧ st[rk] = val], [CSIMPSEQ : ⊦ rk × val ∈ st ↔ ¬ ¬ (rk ∈ st ∧ st[rk] = val)])]]