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Proof step 4 from proof ls-one (
previous step
)
ls(r, a ') = _∃(λ r0. ⌜r0 = null⌝ ∧ r => a × r0) ⊦ ls(r, a ') = r => a × null
User applied rule insert equation
And obtained the following new goals
Proof step 5
:
⊦ _∃(λ r0. ⌜r0 = null⌝ ∧ r => a × r0) = r => a × null
Rule argument
pos: left 1 flags: discard
Additional information