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Proof step 13 from proof subset-put-01 (previous step)

a0 ∈ st → a0 ∈ st0 ∧ st[a0] = st0[a0], a ∈ st → a ∈ st0[a, st[a]] ∧ st[a] = d, a1 ∈ st → a1 ∈ st0[a, d] ∧ st[a1] = st0[a, d][a1], ∀ a0. a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0] ⊦ a0 = a, a ∈ st ∧ st[a] = d ∧ (a0 ∈ st -- a → a0 ∈ st0 -- a ∧ (st -- a)[a0] = (st0 -- a)[a0]), (a1 ∈ st → a1 ∈ st0 ∧ st[a1] = st0[a1]) ∧ ¬ a ∈ st

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: a1 = a

Additional information

Used simplifier rules: Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;%Del-at-strong: a ≠ b ⊦ (st -- b)[a] = st[a];%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%At-other: ⊦ a ≠ b → st[b, d][a] = st[a];%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;%Del-at-strong: a ≠ b ⊦ (st -- b)[a] = st[a]; Used formulas: a1 = a