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Proof step 5 from proof subset-put-strong-05 (previous step)

st ⊆ st0, a2 ∈ st[a, d][a0, d0][a1, d1], a2 ∈ st0[a, d][a0, d0][a1, d1] ⊦ st[a, d][a0, d0][a1, d1][a2] = st0[a, d][a0, d0][a1, d1][a2]

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
st ⊆ st0, a2 ∈ st[a, d][a0, d0][a1, d1], a2 ∈ st0[a, d][a0, d0][a1, d1] ⊦ a2 = a1, st[a, d][a0, d0][a1, d1][a2] = st0[a, d][a0, d0][a1, d1][a2]

Rule argument

fma: a2 = a1

Additional information

Used simplifier rules: At-same: ⊦ st[a, d][a] = d; Used formulas: a2 = a1