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Proof step 2 from proof mod-other-aux1 (previous step)

(a[P] * R)[rs, f, g] ⊦ (a[P] * true)[rs, f, g(r; a0)] ↔ (a[P] * R)[rs, f, g], g(r) = a

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: mod-other-aux: ⊦ g(r) ≠ a ∧ (a[P] * R)[rs, f, g] → (a[P] * true)[rs, f, g(r; a0)];%sep-com: ⊦ R0 * R1 = R1 * R0; from specification heap-sep[heap-aux]%sep-ass: ⊦ (R0 * R1) * R2 = R0 * R1 * R2; from specification heap-sep[heap-aux]