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Proof step 4 from proof ex2-shiftw-inst (
previous step
)
∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ ((_∃ rrP -* P))(H0) ∧ P0(H1) ⊦ (((rrP(r0, r1) -* P) * P0))(H)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 5
:
((_∃ rrP -* P))(H1), P0(H0), H1 ⊥ H0 ⊦ (((rrP(r0, r1) -* P) * P0))(H1 ∪ H0)
Rule argument
Additional information