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Proof step 16 from proof wand-flatten (
previous step
)
P0(H2), P1(H1), H2 ⊥ H1, H ⊥ H2, H ⊥ H1, H1 ⊥ H2 ∪ H ∧ P1(H1) → P2(H1 ∪ H2 ∪ H) ⊦ P2((H2 ∪ H1) ∪ H)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 17
:
P0(H2), P1(H1), H ⊥ H1, H ⊥ H2, H2 ⊥ H1, H1 ⊥ H2 ∪ H → P2(H1 ∪ H2 ∪ H) ⊦ P2((H2 ∪ H1) ∪ H)
Rule argument
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