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Proof step 5 from proof ls-unique (previous step)

(o[ls(r, [])] * R)[rs, cf, af], (o[ls(r, x0)] * R)[rs, cf, af] ⊦ [] = x0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : ls-base, , , ⊦ ls(r, []) = (emp ∧ ⌜r = null⌝ )], [CSIMPSEQ : ⊦ R0 * R1 = R1 * R0], [CSIMPSEQ : ⊦ (R0 * R1) * R2 = R0 * R1 * R2], [CSIMPSEQ : ⊦ emp * R = R], [CSIMPSEQ : ⊦ (⌜bv⌝ ∧ R0) * R1 = (⌜bv⌝ ∧ R0 * R1)], [CSIMPSEQ : ⊦ (P0 ∧ P1) = (P1 ∧ P0)], [CSIMPSEQ : ⊦ ((P0 ∧ P1) ∧ P2) = (P0 ∧ P1 ∧ P2)], [CSIMPSEQ : ⊦ (R0 ∧ R1) = (R1 ∧ R0)], [CSIMPSEQ : ⊦ ((R0 ∧ R1) ∧ R2) = (R0 ∧ R1 ∧ R2)], [CSIMPSEQ : ⊦ (o[⌜bv⌝ ] ∧ emp) = (⌜bv⌝ ∧ emp)], [CSIMPSEQ : ⊦ o[emp] = emp], [CSIMPSEQ : ⊦ o[P0 ∧ P1] = (o[P0] ∧ o[P1])])]]