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Proof step 7 from proof subset-put (previous step)

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

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;%Del-notin: ⊦ ¬ a ∈ st -- a;%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%At-other: ⊦ a ≠ b → st[b, d][a] = st[a];%Del-at-strong: a ≠ b ⊦ (st -- b)[a] = st[a];%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; Used formulas: a0 = a