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Proof step 11 from proof oplus-shift-put-l (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ a0 = a, a0 ∈ st0, (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: in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st); from specification store