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Proof step 11 from proof inj-put (previous step)

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

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: a0 = a

Additional information

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