y ↦ z } cannot have an inverse, since e.g. (x+y) { x ↦ z, y ↦ z } = z+z, and the latter term cannot be transformed back to x+y, as the information about the origin a z stems from is lost. The ground substitution { x ↦ 2 } cannot have an inverse due to a similar loss of origin information e.g. in (x+2) { x ↦ 2 } = 2+2, even if replacing constants by variables was allowed by some fictitious kind of "generalized substitutions". Two substitutions are considered equal if they map each variable to structurally equal result terms, formally: σ = τ if xσ = xτ for each variable x ∈ V. The composition of two substitutions σ = { x1 ↦ t1, …, xk ↦ tk } and τ = { y1 ↦ u1, …, yl ↦ ul } is obtained by removing from the substitution { x1 ↦ t1τ, …, xk ↦ tkτ, y1 ↦ u1, …, yl ↦ ul } those pairs yi ↦ ui for which yi ∈ { x1, …, xk }. The composition of σ and τ is denoted by στ. Composition is an associative operation, and is compatible with substitution application, i.e. (ρσ)τ = ρ(στ), and (tσ)τ = t(στ), respectively, for every substitutions ρ, σ, τ, and every term t. The identity substitution, which maps every variable to itself, is the neutral element of substitution composition. A substitution σ is called idempotent if σσ = σ, and hence tσσ = tσ for every term t. The substitution { x1 ↦ t1, …, xk ↦ tk } is idempotent if and only if none of the variables xi occurs in any ti. Substitution composition is not commutative, that is, στ may be different from τσ, even if σ and τ are idempotent. For example, { x ↦ 2, y ↦ 3+4 } is equal to { y ↦ 3+4, x ↦ 2 }, but different from { x ↦ 2, y ↦ 7 }. The substitution { x ↦ y+y } is idempotent, e.g. ((x+y) {x↦y+y}) {x↦y+y} = ((y+y)+y) {x↦y+y} = (y+y)+y, while the substitution { x ↦ x+y } is non-idempotent, e.g. ((x+y) {x↦x+y}) {x↦x+y} = ((x+y)+y) {x↦x+y} = ((x+y)+y)+y. An example for non-commuting substitutions is { x ↦ y } { y ↦ z } = { x ↦ z, y ↦ z }, but { y