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Proof step 7 from proof mod-other-aux (conclusion)

⊦ g(r) ≠ a ∧ (a[P] * R)[rs, f, g] → (¬ (a[P] * true)[rs, f, g(r ; a0)] ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(a[P] * R)[rs, f, g] ⊦ g(r) = a, (a[P] * true)[rs, f, g(r ; a0)]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ R0 * R1 = R1 * R0], [CSIMPSEQ : ⊦ (R0 * R1) * R2 = R0 * R1 * R2])]]