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Proof step 5 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 6:
a ∈ st0[a, d] ⊕ st1 ⊦ (st0[a, d] ⊕ st1)[a] = d

Rule argument

Additional information

Used simplifier rules: oplus-l-in: ⊦ a ∈ st0 ⊕ st1 ↔ a ∈ st0 ∨ a ∈ st1;%At-same: ⊦ st[a, d][a] = d; from specification store%insin: ⊦ a ∈ st[a, d]; from specification store