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Proof step 6 from proof union-move-left (previous step)

a ∈ st1 ⊦ a ∈ st0, (a0 ∈ st0[a, st1[a]] ∪ (st1 -- a) ↔ a0 ∈ st0 ∪ st1) ∧ (a0 ∈ st0[a, st1[a]] ⊃ st0[a, st1[a]][a0];st1[a0]) = (a0 ∈ st0 ⊃ st0[a0];st1[a0])

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 7:
a ∈ st1 ⊦ a0 = a, a ∈ st0, (a0 ∈ st0[a, st1[a]] ∪ (st1 -- a) ↔ a0 ∈ st0 ∪ st1) ∧ (a0 ∈ st0[a, st1[a]] ⊃ st0[a, st1[a]][a0];st1[a0]) = (a0 ∈ st0 ⊃ st0[a0];st1[a0])

Rule argument

fma: a0 = a

Additional information

Used simplifier rules: union-in: ⊦ a ∈ st0 ∪ st1 ↔ a ∈ st0 ∨ a ∈ st1;% ⊦ a ∈ st[a, d]; (s)% ⊦ st[a, d][a] = d; (s) Used formulas: a0 = a