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Proof step 4 from proof oplus-shift-put-l (previous step)

⊦ (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 weak cut applied rule cut formula

And obtained the following new goals

Proof step 5:
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])
Proof step 8:
⊦ 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])

Rule argument

fma: a0 = a

Additional information

Used formulas: a0 = a