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Proof step 12 from proof eq-shift-del (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 13:
a ∈ st0, ∀ a0. (a0 ∈ st ↔ a0 ∈ st0 -- a) ∧ (a0 ∈ st → st[a0] = (st0 -- a)[a0]) ⊦ a0 = a, a ∈ st, (a0 ∈ st[a, st0[a]] ↔ a0 ∈ st0) ∧ (a0 ∈ st[a, st0[a]] → st[a, st0[a]][a0] = st0[a0])

Rule argument

fma: a0 = a

Additional information

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