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Proof step 8 from proof oplus-shift-put-l (previous step)

⊦ a0 = a, (a0 ∈ st0[a, d] ⊕ st1 ↔ a0 ∈ (st0 ⊕ st1)[a, d]) ∧ (a0 ∈ st0[a, d] ⊕ st1 → (st0[a, d] ⊕ st1)[a0] = (st0 ⊕ st1)[a, d][a0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ a0 = a, (a0 ∈ st0[a, d] ⊕ st1 ↔ a0 ∈ st0 ⊕ st1) ∧ (a0 ∈ st0[a, d] ⊕ st1 → (st0[a, d] ⊕ st1)[a0] = (st0 ⊕ st1)[a0])

Rule argument

Additional information

Used simplifier rules: At-other: ⊦ a ≠ b → st[b, d][a] = st[a]; from specification store%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st); from specification store