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Proof step 6 from proof subset-put-strong-04 (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 7:
a1 ∈ st0[a, d][a0, d0], a1 ∈ st[a, d][a0, d0], st ⊆ st0 ⊦ st[a, d][a0, d0][a1] = st0[a, d][a0, d0][a1], a1 = a0

Rule argument

fma: a1 = a0

Additional information

Used simplifier rules: At-same: ⊦ st[a, d][a] = d; Used formulas: a1 = a0