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Proof step 3 from proof Data-notin-other-2 (previous step)

a0 ∈ st[a, d1][b, d2] ⊦ st[a, d1][b, d2][a0] = d2, st[a, d1][b, d2][a0] = d1, ∃ a1. a1 ∈ st ∧ st[a1] = st[a, d1][b, d2][a0]

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a0 ∈ st[a, d1][b, d2] ⊦ st[a, d1][b, d2][a0] = d1, st[a, d1][b, d2][a0] = d2, a0 = b, ∃ a1. a1 ∈ st ∧ st[a1] = st[a, d1][b, d2][a0]

Rule argument

fma: a0 = b

Additional information

Used simplifier rules: ⊦ st[a, d][a] = d; (s) Used formulas: a0 = b