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Proof step 18 from proof eq-shift-put (previous step)

a ∈ st0, a ∈ st, ∀ a0. (a0 ∈ st ↔ a0 ∈ st0[a, st[a]]) ∧ st[a0] = st0[a, st[a]][a0] ⊦ (a0 ∈ st[a, st0[a]] ↔ a0 ∈ st0) ∧ st[a, st0[a]][a0] = st0[a0]

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 19:
a ∈ st0, a ∈ st, ∀ a0. (a0 ∈ st ↔ a0 ∈ st0[a, st[a]]) ∧ st[a0] = st0[a, st[a]][a0] ⊦ a0 = a, (a0 ∈ st[a, st0[a]] ↔ a0 ∈ st0) ∧ st[a, st0[a]][a0] = st0[a0]

Rule argument

fma: a0 = a

Additional information

Used simplifier rules: notmemlookup: ⊦ ¬ a ∈ st → st[a] = (∅)[a];%At-same: ⊦ st[a, d][a] = d;%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st; Used formulas: a0 = a