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Proof step 4 from proof ex-lst-shift (
previous step
)
∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ o[_∃ xv][rs0, cf, af] ∧ R[rs1, cf, af] ⊦ ∃ x. (o[xv(x)] * R)[rs, cf, af]
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 5
:
rs1 ∩ rs0 = ∅, o[_∃ xv][rs1, cf, af], R[rs0, cf, af] ⊦ ∃ x. (o[xv(x)] * R)[rs1 ∪ rs0, cf, af]
Rule argument
Additional information