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Proof step 16 from proof union-in-01 (previous step)

(∃ sk. r -- sk ∈ st) ∨ (∃ sk. r -- sk ∈ st0) ⊦ ∃ sk. r -- sk ∈ st ∪ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ rk ∈ st ∪ st0 ↔ rk ∈ st ∨ rk ∈ st0
Proof step 15:
r -- sk ∈ st ∨ r -- sk0 ∈ st0 ⊦ ∃ sk1. r -- sk1 ∈ st ∨ r -- sk1 ∈ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]