A question in substitution in logic.

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SUMMARY

The discussion focuses on proving a logical substitution theorem involving the function sub(t;a,b). Specifically, it establishes that if x is the first element of sub(φ;a,ψ), then there exist indices and first elements φ' and ψ' such that x equals sub(φ';a,ψ)ψ'. The user outlines their approach to the proof but seeks further guidance on progressing from their current deductions. The conversation also touches on a separate logical proof regarding implications without using material conditionals.

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  • Understanding of logical substitution functions, specifically sub(t;a,b).
  • Familiarity with logical expressions and their components.
  • Knowledge of implications in propositional logic.
  • Basic skills in formal proof techniques.
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  • Study the properties of logical substitution functions in detail.
  • Research formal proof techniques in propositional logic.
  • Explore the concept of first elements in logical expressions.
  • Learn about implications in logic without using material conditionals.
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Logicians, mathematicians, and students of formal logic who are interested in understanding substitution theorems and implications in propositional logic.

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i need to prove that if x is the first of sub([tex]\phi[/tex];a,[tex]\psi[/tex]) then there exists 1<=i<=n and there exist firsts [tex]\phi' of \phi_i and \psi' for \psi[/tex] such that x=sub([tex]\phi';a,\psi[/tex])[tex]\psi'[/tex]

where sub(t;a,b) is defined as follows:
let a1,..,an be n signs and b1,..,bn expressions.
a=(a1,...,an)
b=(b1,..,bn)
the substitution sub(t;a,b) is defined as:
if t is t1,...,tk then sub(t;a,b)=x1x2...xk
when 1<=i<=k xi is bj if ti=aj and xi is ti when ti isn't in {a1,...,an}.

what i did is as follows:
x is the first of sub(t;a,b) then there exists y such that sub(t;a,b)=xy
and let t' be the first of t, then t=t'z and thuse we can deduce that:
sub(t;a,b)=sub(t';a,b)sub(z;a,b)=xy
but i don't know how to procceed from here.
thanks in davance.
 
Last edited:
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i have another question:
let a|=a' and b|=b' then prove that a'->b|=a->b' without using ~a'|=~a and metrial conditional, i.e without using ~PvQ=P->Q.
 

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