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Proof step 9 from proof ls-unfold1 (previous step)

⊦ r = null, (ls(r, x) * P)(H) ↔ (∃ a, y, r0. (r => a × r0 * ls(r0, y) * P)(H) ∧ x = a ' + y)

User applied rule constructor cut

And closed the goal!

Rule argument

[var-arg: x]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : ls-base, , , ⊦ ls(r, []) = (emp ∧ ⌜r = null⌝ )], [CSIMPGEN : list generated by [], +;], [CSIMPSEQ : ⊦ ⌜false⌝ = false], [CSIMPSEQ : ⊦ (false ∧ P) = false], [CSIMPSEQ : ⊦ ((P0 ∧ P1) ∧ P2) = (P0 ∧ P1 ∧ P2)], [CSIMPSEQ : ⊦ (P0 ∧ P1) = (P1 ∧ P0)], [CSIMPSEQ : ⊦ false * P = false], [CSIMPSEQ : ⊦ (P0 * P1) * P2 = P0 * P1 * P2], [CSIMPSEQ : ⊦ P0 * P1 = P1 * P0], [CSIMPSEQ : ⊦ ¬ false(H)], [CSIMPSEQ : ⊦ a ' + x ≠ []], [CSIMPSEQ : ⊦ (x + y) + z = x + y + z])]]