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Proof step 14 from proof eq-shift-put (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 15:
a ∈ st, (a ∈ st[a, d] ↔ a ∈ st0) ∧ st[a, d][a] = st0[a], ∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ st[a, d][a0] = st0[a0] ⊦ a0 = a, (a0 ∈ st ↔ a0 ∈ st0) ∧ st[a0] = st0[a0] ∧ a ∈ st0 ∧ st0[a] = d

Rule argument

pos: left 2 terms: a abort: F discard: F computed: F

Additional information

Used terms: a