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Proof step 4 from proof sep-split-interactive (
previous step
)
∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P(H0) ∧ P0(H1) ⊦ ∃ H0, H1. H0 ⊥ H1 ∧ H0 ∪ H1 = H ∧ P(H0) ∧ P0(H1)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 5
:
P(H1), P0(H0), H1 ⊥ H0 ⊦ ∃ H3, H2. H3 ⊥ H2 ∧ H3 ∪ H2 = H1 ∪ H0 ∧ P(H3) ∧ P0(H2)
Rule argument
Additional information