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Proof step 13 from proof del-lists-1 (
previous step
)
∃ x. n = # x ∧ ls(r, x)(H) ⊦ r = null, ∃ x. ls(r, x)(H) ∧ n = # x
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 14
:
ls(r, x)(H) ⊦ r = null, ∃ x0. ls(r, x0)(H) ∧ # x = # x0
Rule argument
Additional information