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Proof step 3 from proof map-in (previous step)

∀ i, intfun. i ∈ mapFun(intfun, ints0) ↔ (∃ j. j ∈ ints0 ∧ intfun(j) = i) ⊦ i = intfun(i0) ∨ i ∈ mapFun(intfun, ints0) ↔ (∃ j. (j = i0 ∨ j ∈ ints0) ∧ intfun(j) = i)

User applied rule all left

And obtained the following new goals

Proof step 4:
i ∈ mapFun(intfun, ints0) ↔ (∃ j. j ∈ ints0 ∧ intfun(j) = i), ∀ i, intfun. i ∈ mapFun(intfun, ints0) ↔ (∃ j. j ∈ ints0 ∧ intfun(j) = i) ⊦ i = intfun(i0) ∨ i ∈ mapFun(intfun, ints0) ↔ (∃ j. (j = i0 ∨ j ∈ ints0) ∧ intfun(j) = i)

Rule argument

pos: left 1 terms: i, intfun abort: F discard: F computed: T

Additional information

Used terms: i, intfun