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Proof step 7 from proof Del-in (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 8:
a ∈ st0 -- b ↔ a ≠ b ∧ a ∈ st0 ⊦ a ∈ (st0 -- b)[a0, d] ↔ a ≠ b ∧ a ∈ st0[a0, d], b = a0

Rule argument

pos: left 1 terms: a, b abort: F discard: T computed: T

Additional information

Used terms: a, b