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Proof step 11 from proof embed-and (previous step)

⊦ ((P0 ∧ P1))(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a) ↔ a[P0](mk(rs, f, g)) ∧ a[P1](mk(rs, f, g))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ P0(mk(rs, f)) ∧ P1(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a) ↔ a[P0](mk(rs, f, g)) ∧ a[P1](mk(rs, f, g))

Rule argument

Additional information

Used simplifier rules: and-com: ⊦ (P0 ∧ P1) = (P1 ∧ P0); from specification heap-sep%and-ass: ⊦ ((P0 ∧ P1) ∧ P2) = (P0 ∧ P1 ∧ P2); from specification heap-sep% ⊦ ((P0 ∧ P1))(H) ↔ P0(H) ∧ P1(H); (s)