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Proof step 28 from proof union-def (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 1:
⊦ st ∪ ∅ = st
Proof step 27:
∀ rk, st, val. rk ∈ st ∪ st1 ∧ (st ∪ st1)[rk] = val ↔ ¬ rk ∈ st1 ∧ rk ∈ st ∧ st[rk] = val ∨ rk ∈ st1 ∧ st1[rk] = val ⊦ rk ∈ st ∪ st1[rk0, val0] ∧ (st ∪ st1[rk0, val0])[rk] = val ↔ ¬ rk ∈ st1[rk0, val0] ∧ rk ∈ st ∧ st[rk] = val ∨ rk ∈ st1[rk0, val0] ∧ (st1[rk0, val0][rk]) = val

Rule argument

[VARWITHVARSEQSARG : st0, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : store generated by ∅, [];], [CSIMPSEQ : ⊦ ¬ rk ∈ ∅])]]